# determine whether the geometric series `sum_{n=0}^infty e^{-n/2}` converges or diverges. If the series converges find its sum.

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### 2 Answers

A geometric series `sum_{n=0}^infty r^n` converges to `1/{1-r}` if and only if `|r|<1` .

In this case, we have the nth term as

`e^{-n/2}`

`=(e^{-1/2})^n` using the rule `(a^m)^n=a^{mn}`

which means that `r=e^{-1/2}\approx 0.607 <1`

so the series converges and the sum is

`1/{1-r}`

`=1/{1-e^{-1/2}}`

`=e^{1/2}/{e^{1/2}-1}`

**The series converges to the sum `e^{1/2}/{e^{1/2}-1}` .**

**Sources:**

The power series 1+a+a^2+a^3+...+a^n+... ,where = 0.1.2,3.....to inf converges when |a| <1.

The nth tern e^(-0.5n) = (e^-0.5)^n.

We know e is number between 2 and 3.

So (3^-0.5) < e^(-0.5) < (2^-0.5)

=> e^(-0.5) < 1

Therefore 1+(e^-5)+(e^-5)2+(e^-5)^3+....(e^-5)n+....must converge.

So **sum (from n =0 to n= infinity) e^(-0.5n) must converge.**