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Algebra Level 4

$\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^4 + 4}$

If the value of the above summation above is equal to $\dfrac{p}{q}$ , where $p$ and $q$ are coprime positive integers, find $p + q$ .

The answer is 11.

1 solution

Akshat Sharda
Jan 3, 2016

$\begin{aligned} & = \displaystyle \sum^{\infty}_{n=1} \frac{n}{n^4+4} \\ & = \displaystyle \sum^{\infty}_{n=1}\frac{n}{(n^2+2+2n)(n^2+2-2n)} \\ & = \frac{1}{4}\displaystyle \sum^{\infty}_{n=1} \left( \frac{1}{n^2+2-2n}-\frac{1}{n^2+2+2n}\right) \\ & = \frac{1}{4}\left(1-\cancel{\frac{1}{5}}+\frac{1}{2}-\cancel{\frac{1}{10}}+ \cancel{\frac{1}{5}}-\cancel{\frac{1}{17}}+ \cancel{\frac{1}{10}}-\cancel{\frac{1}{26}}+ \cancel{\frac{1}{17}}- \cancel{\frac{1}{37}}+ \cancel{\frac{1}{26}}- \cancel{\frac{1}{50}}+\ldots \right) \\ & = \frac{1}{4}\cdot \left(1+\frac{1}{2}\right) \\ & = \frac{3}{8} \\ \Rightarrow 3+8 &= \boxed{11}\end{aligned}$

Challenge: Can you derive a partial sum formula?

Aditya Agarwal - 5 years, 5 months ago

Your challenge is not clear to me !! What do you want to say ?

Akshat Sharda - 5 years, 5 months ago

Why? Partial sums. You have to derive a formula for the sum of it till $n$ terms.

Aditya Agarwal - 5 years, 5 months ago

@Aditya Agarwal – $= \frac{1}{4}\displaystyle \sum^{k}_{n=1} \left( \frac{1}{n^2+2-2n}-\frac{1}{n^2+2+2n}\right)$

Akshat Sharda - 5 years, 5 months ago

@Akshat Sharda – (In response to previous two comments). It works and is a nice solution, but suppose instead of $\infty$ , there is any number, like $1000$ , then? Then you have to derive a general formula. Like $\sum\limits^\infty_{k=1}k=\text{Divergent}$ , but $\sum\limits^{n}_{k=1}k=\frac{n^2+n}{2}$ . (This is a partial sum)

Aditya Agarwal - 5 years, 5 months ago

@Aditya Agarwal – $=\huge \frac{n(n+1)(3n^2+3n+2) }{8(n^2+1)(n^2+2n+2) }$

Akshat Sharda - 5 years, 5 months ago

@Akshat Sharda – (y) Great!

Aditya Agarwal - 5 years, 5 months ago

@Aditya Agarwal – Doesn't this work ? Or I should get a formula more easier to compute.

Akshat Sharda - 5 years, 5 months ago

Seen it before

The answer is 11.

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