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  1. Sum the series to infinite terms: $$1+\frac14+\frac{1\cdot3}{4\cdot8}+\frac{1\cdot3\cdot5}{4\cdot8\cdot12}+...$$

I couldn't get any clue on how to solve this summation. I tried to solve without any help and the things I got confused with : [...]

Is how to, do summation on the general term, that's: $$m=\sum_{n=0}^\infty\frac{(2n+1)!}{2^n\cdot n!\cdot 4\cdot n!}$$

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    Please improve the question by adding your own effort. – 2017-01-29
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    Thanks for your suggestion. – 2017-01-29
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    The denominator is $6 \cdot n!$, not $2^n \cdot n!$ Your equation is also wrong because the left side depends on $n$ and the right does not. – 2017-01-29
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    Man a huge mistake ... So sorry – 2017-01-29
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    Sir I had typed a wrong question please forgive me. – 2017-01-29
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    The numerator is not $(2n+1)!$ because it is missing the even terms. You mean $(2n+1)!!$ The denominator is now $4 \cdot n!$, not $2^n \cdot n!$ – 2017-01-29

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The generalized binomial theorem tells you $$(1-z)^{-1/2}= \sum_{n=0}^{\infty} \Big( \frac{1}{2} \cdot \frac{3}{2} \cdot ... \cdot \frac{2n-1}{2}\Big) \frac{z^n}{n!}$$ $$= 1 + \frac{1}{2}z + \frac{1 \cdot 3}{2 \cdot 4}z^2 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} z^3 + ...$$

and evaluating at $z=1/2$ gives you the sum $\sqrt{2}.$