$c_n = ∑a_k a_n-k$ is convergent ot not

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Let $a_n =frac (-1)^n(1+n)^0.5$ and let $c_n = ∑a_k a_n-k$



Then will $sum^infty _n=0 c_n$ be converge?



If $sum^infty _n=0 a_n$ converged absolutely I could say that the sequence $sum^infty _n=0 c_n$ .



But in this case I have no idea how to solve it.






real-analysis sequences-and-series cauchy-sequences







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  • i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
    – sajan
    Sep 8 at 6:51















up vote
1
down vote

favorite












Let $a_n =frac (-1)^n(1+n)^0.5$ and let $c_n = ∑a_k a_n-k$



Then will $sum^infty _n=0 c_n$ be converge?



If $sum^infty _n=0 a_n$ converged absolutely I could say that the sequence $sum^infty _n=0 c_n$ .



But in this case I have no idea how to solve it.






real-analysis sequences-and-series cauchy-sequences







share|cite|improve this question





















  • i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
    – sajan
    Sep 8 at 6:51













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $a_n =frac (-1)^n(1+n)^0.5$ and let $c_n = ∑a_k a_n-k$



Then will $sum^infty _n=0 c_n$ be converge?



If $sum^infty _n=0 a_n$ converged absolutely I could say that the sequence $sum^infty _n=0 c_n$ .



But in this case I have no idea how to solve it.






real-analysis sequences-and-series cauchy-sequences







share|cite|improve this question













Let $a_n =frac (-1)^n(1+n)^0.5$ and let $c_n = ∑a_k a_n-k$



Then will $sum^infty _n=0 c_n$ be converge?



If $sum^infty _n=0 a_n$ converged absolutely I could say that the sequence $sum^infty _n=0 c_n$ .



But in this case I have no idea how to solve it.






real-analysis sequences-and-series cauchy-sequences




real-analysis sequences-and-series cauchy-sequences






share|cite|improve this question













share|cite|improve this question











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asked Sep 8 at 6:37









cmi

946110




946110











  • i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
    – sajan
    Sep 8 at 6:51

















  • i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
    – sajan
    Sep 8 at 6:51
















i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
– sajan
Sep 8 at 6:51





i think in case of $c_n = ∑a_k a_n-k$, $ n ge k$. and as $a_k$ some constant it doesn make any change to nature of series.
– sajan
Sep 8 at 6:51











1 Answer
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We have that



$$c_n = sum_k=0^n a_k a_n-k=c_n = sum_k=0^n frac (-1)^k(-1)^n-k(1+k)^0.5(1+n-k)^0.5= sum_k=0^n frac (-1)^n(1+n+nk-k^2)^0.5not to 0$$



indeed for $n$ even



$$sum_j=-frac n2^frac n2 frac 1(1+n+n(j+n/2)-(j+n/2)^2)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+nj+n^2/2-j^2-nj-n^2/4)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+n^2/4-j^2)^0.5=2sum_j=0^frac n2 frac 1(1+n+n^2/4-j^2)^0.5neq 0$$



and therefore $sum^infty _n=0 c_n$ does not converge.






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    We have that



    $$c_n = sum_k=0^n a_k a_n-k=c_n = sum_k=0^n frac (-1)^k(-1)^n-k(1+k)^0.5(1+n-k)^0.5= sum_k=0^n frac (-1)^n(1+n+nk-k^2)^0.5not to 0$$



    indeed for $n$ even



    $$sum_j=-frac n2^frac n2 frac 1(1+n+n(j+n/2)-(j+n/2)^2)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+nj+n^2/2-j^2-nj-n^2/4)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+n^2/4-j^2)^0.5=2sum_j=0^frac n2 frac 1(1+n+n^2/4-j^2)^0.5neq 0$$



    and therefore $sum^infty _n=0 c_n$ does not converge.






    share|cite|improve this answer
























      up vote
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      down vote













      We have that



      $$c_n = sum_k=0^n a_k a_n-k=c_n = sum_k=0^n frac (-1)^k(-1)^n-k(1+k)^0.5(1+n-k)^0.5= sum_k=0^n frac (-1)^n(1+n+nk-k^2)^0.5not to 0$$



      indeed for $n$ even



      $$sum_j=-frac n2^frac n2 frac 1(1+n+n(j+n/2)-(j+n/2)^2)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+nj+n^2/2-j^2-nj-n^2/4)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+n^2/4-j^2)^0.5=2sum_j=0^frac n2 frac 1(1+n+n^2/4-j^2)^0.5neq 0$$



      and therefore $sum^infty _n=0 c_n$ does not converge.






      share|cite|improve this answer






















        up vote
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        down vote










        up vote
        0
        down vote









        We have that



        $$c_n = sum_k=0^n a_k a_n-k=c_n = sum_k=0^n frac (-1)^k(-1)^n-k(1+k)^0.5(1+n-k)^0.5= sum_k=0^n frac (-1)^n(1+n+nk-k^2)^0.5not to 0$$



        indeed for $n$ even



        $$sum_j=-frac n2^frac n2 frac 1(1+n+n(j+n/2)-(j+n/2)^2)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+nj+n^2/2-j^2-nj-n^2/4)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+n^2/4-j^2)^0.5=2sum_j=0^frac n2 frac 1(1+n+n^2/4-j^2)^0.5neq 0$$



        and therefore $sum^infty _n=0 c_n$ does not converge.






        share|cite|improve this answer












        We have that



        $$c_n = sum_k=0^n a_k a_n-k=c_n = sum_k=0^n frac (-1)^k(-1)^n-k(1+k)^0.5(1+n-k)^0.5= sum_k=0^n frac (-1)^n(1+n+nk-k^2)^0.5not to 0$$



        indeed for $n$ even



        $$sum_j=-frac n2^frac n2 frac 1(1+n+n(j+n/2)-(j+n/2)^2)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+nj+n^2/2-j^2-nj-n^2/4)^0.5=sum_j=-frac n2^frac n2 frac 1(1+n+n^2/4-j^2)^0.5=2sum_j=0^frac n2 frac 1(1+n+n^2/4-j^2)^0.5neq 0$$



        and therefore $sum^infty _n=0 c_n$ does not converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 8 at 7:20









        gimusi

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