# why $\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{-2} - 3$ [closed]

please how to show it ? I need to find the exact sum of :
$$\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{-2} - 3$$

## closed as off-topic by José Carlos Santos, Scientifica, Theoretical Economist, Henning Makholm, rtybaseSep 23 '18 at 23:15

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• Welcome to MSE. Question like this stating "Here is my task. Solve it for me" are poorly received one this website. Show what you have done so far and where exactly you are struggling and other users are more likely to help you :) – mrtaurho Sep 23 '18 at 18:52
• yes but it's just to learn because it's to determine some matrix exponential – KEVIN DLL Sep 23 '18 at 18:53
• Are you familiar with the Taylor series expansion of the exponential? – mrtaurho Sep 23 '18 at 18:54
• The equality is not true. It'll be $e^2-3$ in the right hand side. Use the series expansion $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ for $x=2$ to prove that. – Surajit Sep 23 '18 at 18:56
• Because your sum starts from $n=2$ and not from $n=0$ like in the taylor series of the exponential. – Mark Sep 23 '18 at 19:00

Consider the Taylor series expansion $$e^x=\sum_{i=0}^{\infty}\frac{x^n}{n!}.$$

So at $$x=2,\quad e^2=\sum_{n=0}^{\infty}\frac{2^n}{n!}=3+\sum_{n=2}^{\infty}\frac{2^n}{n!}\implies \sum_{n=2}^{\infty}\frac{2^n}{n!}=e^2-3.$$

You wanted $$\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{2} - 3$$

You need to know the Taylor series for $$e^x$$, namely

$$e^x = \sum_{n=0}^{+\infty}\frac{(x)^n}{n!}=1+x+\sum_{n=2}^{+\infty}\frac{(x)^n}{n!}$$

All you have to do from here is to let $$x=2$$ and subtract the first two terms.

HINT

Try to use the Taylor series expansion of the exponential which is given by

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\sum_{n=2}^{\infty}\frac{x^n}{n!}$$

Hint: $$a+b = c+d$$ implies $$b = c+d -a$$ Can you take it from here?

Note that

$$\sum_{n=2}^{\infty}\frac{(2)^n}{n!} >\sum_{n=2}^{2}\frac{(2)^n}{n!} = 2> e^{-2} -3$$

therefore the given equality cannot be true.