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I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$ I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.

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You are almost there.

So once you have obtained $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ then observe the following:

$n^2+2n+2=(n+1)^2+1$ and $n^2-2n+2=(n-1)^2+1$. Which means the given sum becomes $$\sum\limits_{n=2}^\infty \left[\frac{-1}{2^n\{(n+1)^2+1\}}+\frac{1}{2^{n-2}\{(n-1)^2+1\}} \right] +\sum\limits_{n=2}^\infty \frac{1}{2^n}=:\sum\limits_{n=2}^\infty [u_n-u_{n+2}]+\frac 12 $$ where $u_n:=\frac{1}{2^{n-2}\{(n-1)^2+1\}}$.

If you calculate $\sum\limits_{n=2}^m[u_n-n_{n+2}]$, you will find $u_2+u_3-(u_{m+1}+u_{m+2})$. hence taking $m\rightarrow \infty$ we shall get $$\sum\limits_{n=2}^\infty [u_n-u_{n+2}]=u_2+u_3=\frac{1}{2}+\frac{1}{10}.$$

Hence the final sum is $2\times \frac 12+\frac{1}{10}=\cdots.$

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  • $\begingroup$ (+1) Who does not love telescoping magic? $\endgroup$ – Jack D'Aurizio Feb 9 '15 at 10:52
  • $\begingroup$ Ya I realized after I posted my answer. Edited just now. Thanks for the alert $\endgroup$ – Anjan3 Feb 9 '15 at 10:53
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Note that $$\dfrac{n^4+3n^2+10n+10}{2^n(n^4+4)}=\dfrac{1}{2^n}+\dfrac{3n^2+10n+6}{2^n[(n^2+2)^2-(2n)^2]}$$ Then let's find constants $A,B$ suct that $$\dfrac{3n^2+10n+6}{(n^2+2n+2)(n^2-2n+2)}=\dfrac{A(n+1)+B}{(n+1)^2+1}-4\Big[\dfrac{A(n-1)+B}{(n-1)^2+1}\Big]$$ to obtain the form $$f(n+1)-f(n-1).$$ For $n=-1,$ we have $-\dfrac{1}{5}=B+4\Big(\dfrac{2A-B}{5}\Big)\iff8A+B=-1.$

For $n=+1,$ we have $\dfrac{19}{5}=\Big(\dfrac{2A+B}{5}\Big)-4B\iff2A-19B=19.$
By solving these equations, $$A=0,\,\,\,\,\,B=-1$$
Now $$\dfrac{n^4+3n^2+10n+10}{2^n(n^4+4)}=\dfrac{1}{2^n}-\dfrac{1}{2^n((n+1)^2+1)}+\dfrac{1}{2^{n-2}((n-1)^2+1)}$$ Can you continue from here? Good Luck.

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