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Given set $A=\{ \frac{i^2+1}{2}\ |\ i\in\mathbb{N},\ i\geq 65,\ 2\nmid i \}$ calculate sum of reciprocals of values of its elements.

So first I simplified the set:

$$ A=\{ \frac{(2j+1)^2+1}{2}\ |\ j\in\mathbb{N},\ j\geq 32 \}=\{ 2j^2+2j+1\ |\ j\in\mathbb{N},\ j\geq 32 \} $$

And now the sum we are looking can be calculated with the expression: $$ \sum_{j=32}^{\infty}\frac{1}{2j^2+2j+1}=\sum_{j=0}^{\infty}\frac{1}{2j^2+2j+1}-\sum_{j=0}^{31}\frac{1}{2j^2+2j+1} $$

However I'm not sure how to carry on. Can you help me solve it?

MartinYakuza
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  • Do you want an approximation (use Euler-Maclaurin summation for example) or an exact value (very unlikely)? – Greg Martin Aug 28 '20 at 16:54
  • @GregMartin I think the first part (0 to infinity) of this should an exact formula, and the second part (0 to 31) can be approximated (unless you can give me exact value to both). – MartinYakuza Aug 28 '20 at 16:59
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    WolframAlpha says that the first sum equals $\frac{1}{2} \pi \tanh \left(\frac{\pi }{2}\right)$, and the second sum is just the sum of finitely many rational numbers and can be worked out exactly. – Greg Martin Aug 28 '20 at 17:02
  • @GregMartin ok, so just help with the sum from 0 to infinity – MartinYakuza Aug 28 '20 at 17:05
  • It looks like some complex analysis is typically involved in evaluating the main sum. Here's a related question that may help: https://math.stackexchange.com/questions/736860/find-the-infinite-sum-of-the-series-sum-n-1-infty-frac1n2-1 – Jason Aug 30 '20 at 03:30

1 Answers1

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Lets start with: $$\eqalign{ \sum_{j\in\mathbb{Z}}\frac{1}{2j^2+2j+1}&=\color{red}{\sum_{- \infty \le j \le -1}\frac{1}{2j^2+2j+1}}+\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr &=\color{red}{\sum_{\infty \ge -j \ge 1}\frac{1}{2j^2+2j+1}} +\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr &=\color{red}{\sum_{\infty \ge \underbrace{-j-1}_{-j-1:=k} \ge 0}\frac{1}{2j^2+2j+1}} +\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr &=\color{red}{\sum_{0 \le k \le \infty }\frac{1}{2(-k-1)^2+2(-k-1)+1}} +\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr &=\color{red}{\sum_{0 \le k \le \infty }\frac{1}{2k^2+2k+1}} +\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr &=2\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}\cr }$$

therefore:

$$\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}=\frac{1}{2}\sum_{j\in\mathbb{Z}}\frac{1}{2j^2+2j+1}$$

due to Infinite Series and the Residue Theorem black sum it is easy to count:

$$\eqalign{\color{blue}{\sum_{0 \le j \le \infty }\frac{1}{2j^2+2j+1}}&= -\frac{1}{2}\left( \text{Res}\left\{\frac{ \pi \cot \left( \pi z\right) }{2z^2+2z+1} , \frac{-1-i}{2} \right\} +\text{Res}\left\{\frac{ \pi \cot \left( \pi z\right) }{2z^2+2z+1} ,\frac{-1+i}{2} \right\} \right)\cr &=\frac{ \pi }{2}\text{tgh}\left( \frac{ \pi }{2} \right)}$$

summarizing with the litle help of Wolfram for finite sum

$$\sum_{32 \le j \le \infty }\frac{1}{2j^2+2j+1}=\\ \frac{ \pi }{2}\text{tgh}\left(\frac{ \pi }{2} \right)-\frac{11331201497882268207659413442681772956711855668184103152864}{7951513543813219897041288425928485300865169239957975379625}$$

Marek Kryspin
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