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We had integrals, so we have to have series as well. Here are 10 easy to difficult series and infinite products. Up to you to find out the exact sum.

**Rules:**- The answer must be a finite expression. The only expressions allowed are integers written in base 10, the elementary arithmetic operations, trigonometric functions and their inverses, exponential functions and their inverses, ##\pi## and imaginary numbers.
- Any use of sources is allowed as long as you don't search for the question directly. So you can search a calculus book, but not type in the question in wolframalpha. Any use of sources must be detailed in the answer.
- The first person to give a complete answer will get the recognition. Sometimes another person will be given the recognition as well.
- If you encountered the question before and remember the answer, you cannot participate in this question.

**Problems:**- SOLVED BY fresh_42 ##\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!} = 1##

- SOLVED BY Math_QED ##\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{4n^2 - 1} = \frac{\pi-2}{4}##

- SOLVED BY fresh_42 ##\sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^6} = \frac{\pi^6}{960}##

- SOLVED BY fresh_42 ##\prod_{n=0}^{+\infty} \left(1 - \frac{4}{\left(n + \frac{1}{2}\right)^2}\right) = 1##

- SOLVED BY Svein ##\sum_{n=1}^{+\infty} 2^{-n} \sin(2n) = \frac{2\sin(2)}{5-4\cos(2)}##

- SOLVED BY Fightfish ##\prod_{k=1}^{+\infty} \cos\left(\frac{1}{2^k}\right) = \sin(1)##

- SOLVED BY Fightfish ##\sum_{n=0}^{+\infty} \frac{(2n)!}{16^n (n!)^2 (2n+1)}=2\sin^{-1}\left(\frac{1}{2}\right)##

- SOLVED BY bpet ##\prod_{n=0}^{+\infty} \sqrt[2^{n+1}]{|\tan(2^n)|}##

- SOLVED BY fresh_42 ##\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}}=\frac{4}{25\sqrt{5}}\pi^2## where ##F_n## is the ##n##th Fibonacci number.

- SOLVED BY geoffrey159 ##\prod_{n=1}^{+\infty} \frac{4n^2}{4n^2 - 1} = \frac{\pi}{2}##

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