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Find this follow sum $$\sum_{n=0}^{\infty}(-1)^{n}\dfrac{n+1}{(2n+1)!}$$

My try:since $$\sum_{n=0}^{\infty}(-1)^n\dfrac{n+1}{(2n+1)!}=\dfrac{1}{2}\sum_{n=0}^{\infty}(-1)^n\dfrac{2n+1+1}{(2n+1)!}=\dfrac{1}{2}\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n)!}+\dfrac{1}{2}\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n+1)!}=\dfrac{1}{2}[\cos{1}+\sin{1}]$$

My question: have other methods? Thank you

1 Answers1

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$$e^y=\sum_{r=1}^{\infty}\frac{y^r}{r!}$$

and $$e^{ix}=\cos x+i\sin x,e^{-ix}=\cos x-i\sin x\implies e^{ix}\pm e^{-ix}=?$$

Now, $$\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n)!}=\sum_{n=0}^{\infty}\dfrac{i^{2n}}{(2n)!}=2(e^i+e^{-i})$$

$$\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n+1)!}=\frac1i\sum_{n=0}^{\infty}i^{2n+1}\dfrac{1}{(2n+1)!}=\frac{2(e^i-e^{-i})}i=-2i(e^i-e^{-i})$$

lab bhattacharjee
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