What value does

$$\sum_{n=1}^{\infty} \dfrac{1}{4n^2+16n+7}$$ converge to?

Ok so I've tried changing the sum to:

$$\sum_{n=1}^{\infty} \dfrac{1}{6(2n+1)}-\dfrac{1}{6(2n+7)}$$

and then writting some values: $$\frac16·(\frac13+\frac15+\frac17\dots+\frac1{2N+1})-\frac16·(\frac19+\frac1{11}+\frac1{13}\dots+\frac1{2N+7})$$

but I don't know what else I can do to finish it! Any hint or solution?

  • 1
    $\begingroup$ In those sums you’ve written out, everything cancels except finitely many terms. For example, in the first sum, the very next term inside the $\cdots$ is $\frac 1 9$ which cancels with the $\frac 19$ in the second sum. $\endgroup$ – User8128 Dec 8 '18 at 15:36

Hint: Let's look at the $100$th partial sum. It's good to get some concreteness.


We have a bunch of terms that are repeated: $\frac{1}{9}+\cdots+\frac{1}{201}$ exists in each bracketed portion, so we can simply cancel all of them out to get


Can you see how to use this line of reasoning to get the answer?


Check that $$\left(\frac13+\frac15+\frac17+\frac19+\frac1{11}+\cdots+\frac1{2N+1}\right)-$$ $$-\left(\frac19+\frac1{11}+\cdots+\frac1{2N+1}+\frac1{2N+3}+\frac1{2N+5}+\frac1{2N+7}\right)=$$ $$=\frac13+\frac15+\frac17-\frac1{2N+3}-\frac1{2N+5}-\frac1{2N+7}.$$

And if you take the limit as $N\to\infty$ this becomes just $$\frac13+\frac15+\frac17.$$


After given

$$\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)(2n+7)}=\frac1{36}\sum_{n=1}^{\infty} \dfrac{1}{(\frac{n}{3}+\frac{1}{6})(\frac{n}{3}+\frac{7}{6})}$$

set function

$$f(x)=\frac1{36}\sum_{n=1}^{\infty} \dfrac{x^{\frac{n}{3}+\frac{7}{6}}}{(\frac{n}{3}+\frac{1}{6})(\frac{n}{3}+\frac{7}{6})}$$

then take the second derivative of function $f(x)$, which is

$$f''(x)=\frac1{36}\sum_{n=1}^{\infty} x^{\frac{n}{3}-\frac{5}{6}}$$

and it is easy to find this series is equal to


also, notice that you also get the boundary $f'(0)=0$ and $f(0)=0$, then you can do the integral twice to find the original $f(x)$, which is


and take the limitation for $x\to1$ which is the result

$$\sum_{n=1}^{\infty} \dfrac{1}{4n^2+16n+7}= \lim_{x \to 1} f(x)=\frac{71}{630}$$

Actually, the general function method is much more complicated than fraction splitting.


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