- #1

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- 81

- Homework Statement
- Determine if the series ##\sum_{n=1}^{\infty} \frac{2^n}{n !}## converges.

- Relevant Equations
- Partial sums will be denoted by ##s_n##.

##s_1=2##

##s_2=4##

##s_3=5.333##

##s_4=5.9999##

##(s_n)## is increasing, but unable to guess a bound. Let's see if Cauchy criterion can do something.

For n>2,

$$

s_{n+k} - s_n = \frac{2^{n+1} }{(n+1)!} + \frac{ 2^{n+2} }{(n+2)!} + \cdots \frac{2^{n+k} }{(n+k)!}

$$

$$

s_{n+k} - s_n < \frac{2^{n+1} }{ (n+1)!} + \frac{ 2^{n+2} }{n (n+1)!} + \frac{ 2^{n+3} }{n^2(n+1)!} + \cdots \frac{2^{n+k} }{n^{k-1}(n+1)!}

$$

The RHS is a Geometric series with ## a = \frac{2^{n+1} }{ (n+1)!}## and ##r = \frac{2}{n}##. So, by sum formula of Geometric series (as r is less than 1, coz we assumed n >2, therefore series converges)

$$

s_{n+k} - s_n < \frac{2^n}{(n+1)!} \frac{n^k - 2^k}{n^{k-1} } ~\frac{1}{n-2} \lt \frac{2^n}{(n+1)!} \frac{(n-2)~k~n^{k-1}}{n^{k-1}}= \frac{2^n}{(n+1)!} k

$$

Am I correct so far? My doubt is though ##\frac{2^n}{(n+1)!}## can be made as small as we please by taking large n, but won't taking a far off term make ##k## very big, thus, nullifying the nulling effect of ##\frac{2^n}{(n+1)!}##?

##s_2=4##

##s_3=5.333##

##s_4=5.9999##

##(s_n)## is increasing, but unable to guess a bound. Let's see if Cauchy criterion can do something.

For n>2,

$$

s_{n+k} - s_n = \frac{2^{n+1} }{(n+1)!} + \frac{ 2^{n+2} }{(n+2)!} + \cdots \frac{2^{n+k} }{(n+k)!}

$$

$$

s_{n+k} - s_n < \frac{2^{n+1} }{ (n+1)!} + \frac{ 2^{n+2} }{n (n+1)!} + \frac{ 2^{n+3} }{n^2(n+1)!} + \cdots \frac{2^{n+k} }{n^{k-1}(n+1)!}

$$

The RHS is a Geometric series with ## a = \frac{2^{n+1} }{ (n+1)!}## and ##r = \frac{2}{n}##. So, by sum formula of Geometric series (as r is less than 1, coz we assumed n >2, therefore series converges)

$$

s_{n+k} - s_n < \frac{2^n}{(n+1)!} \frac{n^k - 2^k}{n^{k-1} } ~\frac{1}{n-2} \lt \frac{2^n}{(n+1)!} \frac{(n-2)~k~n^{k-1}}{n^{k-1}}= \frac{2^n}{(n+1)!} k

$$

Am I correct so far? My doubt is though ##\frac{2^n}{(n+1)!}## can be made as small as we please by taking large n, but won't taking a far off term make ##k## very big, thus, nullifying the nulling effect of ##\frac{2^n}{(n+1)!}##?