Evaluate the limit of...











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$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







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  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22















up vote
4
down vote

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$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







share|cite|improve this question




















  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22













up vote
4
down vote

favorite
4









up vote
4
down vote

favorite
4






4





$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







share|cite|improve this question















$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 15:31









Asaf Karagila♦

301k32422753




301k32422753










asked Dec 4 at 14:18









user605734 MBS

1789




1789








  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22














  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22








3




3




You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
– xbh
Dec 4 at 14:22




You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
– xbh
Dec 4 at 14:22










3 Answers
3






active

oldest

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up vote
5
down vote



accepted










HINT



We have that by Stolz-Cesaro



$$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






share|cite|improve this answer




























    up vote
    2
    down vote













    Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
    and
    $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
    Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
    ?






    share|cite|improve this answer






























      up vote
      1
      down vote













      Thanks to user xbh for the hint:



      Using the Stolz-Cesàro theorem, we have:



      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



      $b_n=n^2$



      Two monotone and increasing sequences.



      Apply the theorem to get:



      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



      Then:



      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






      share|cite|improve this answer

















      • 3




        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
        – gimusi
        Dec 4 at 14:55













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      HINT



      We have that by Stolz-Cesaro



      $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



      $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






      share|cite|improve this answer

























        up vote
        5
        down vote



        accepted










        HINT



        We have that by Stolz-Cesaro



        $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



        $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






        share|cite|improve this answer























          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          HINT



          We have that by Stolz-Cesaro



          $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



          $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






          share|cite|improve this answer












          HINT



          We have that by Stolz-Cesaro



          $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



          $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 at 14:43









          gimusi

          92.5k94495




          92.5k94495






















              up vote
              2
              down vote













              Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
              and
              $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
              Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
              ?






              share|cite|improve this answer



























                up vote
                2
                down vote













                Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                and
                $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                ?






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                  and
                  $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                  Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                  ?






                  share|cite|improve this answer














                  Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                  and
                  $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                  Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                  ?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 4 at 14:42

























                  answered Dec 4 at 14:36









                  Robert Israel

                  316k23206457




                  316k23206457






















                      up vote
                      1
                      down vote













                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer

















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55

















                      up vote
                      1
                      down vote













                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer

















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55















                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer












                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 4 at 14:48









                      user605734 MBS

                      1789




                      1789








                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55
















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55










                      3




                      3




                      Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                      – gimusi
                      Dec 4 at 14:55






                      Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                      – gimusi
                      Dec 4 at 14:55




















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                      Morgemoulin

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                      Morgemoulin — miejscowość i gmina — Państwo   Francja Region Grand Est Departament Moza Okręg Verdun Kod INSEE 55357 Powierzchnia 6,82 km² Populacja  (1990) • liczba ludności 75 • gęstość 11 os./km² Kod pocztowy 55400 Położenie na mapie Francji Morgemoulin 49°14′N   5°35′E / 49,233333   5,583333 Portal Francja Morgemoulin – miejscowość i gmina we Francji, w regionie Grand Est, w departamencie Moza. Według danych na rok 1990 gminę zamieszkiwało 75 osób, a gęstość zaludnienia wynosiła 11 osób/km² (wśród 2335 gmin Lotaryngii Morgemoulin plasuje się na 970. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 848.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Verdun Abaucourt-Hautecourt • Aincreville • Ambly-sur-Meuse • Amel-sur-l'Étang • Ancemont • Arrancy-sur-Crusne • Aubréville • Autré
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                      Scott Moir

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                      Scott Moir Tessa Virtue i Scott Moir w 2016 r. Reprezentacja   Kanada Data i miejsce urodzenia 2 września 1987 London Wzrost 180 cm Konkurencja Pary taneczne Partner sportowy Tessa Virtue Trener Marie-France Dubreuil, Patrice Lauzon, Romain Haguenauer Poprzednio: Marina Zujewa, Oleg Epstein, Igor Szpilband, Johnny Johns, Carol Moir, Paul MacIntosh, Suzanne Killing Klub Montreal International School of Skating Poprzednio: Arctic Edge FSC Ilderton Skating Club Lokalizacja treningowa Montreal Poprzednio: Canton Kitchener Rekordy życiowe ISU przed sezonem 2018/2019 Nota łączna 206,07 Igrzyska Olimpijskie 2018 Taniec krótki 83,67 Igrzyska Olimpijskie 2018 Taniec dowolny 122,40 Igrzyska Olimpijskie 2018 Dorobek medalowy Reprezentacja   Kanada Igrzyska olimpijskie Złoto Vancouver 2010 pary taneczne Złoto Pjongczang 2018 pary taneczne Złoto Pjongczang 2018 drużynowo Srebro S
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                      Souastre

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                      Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe
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