sequences-and-series - convergence - (Dis-)proving the series $\sum\limits_n\left( 1+ \frac{1}{n} \right)^n$ converges - answerstu - answerstu.com answerstu

convergence - (Dis-)proving the series $\sum\limits_n\left( 1+ \frac{1}{n} \right)^n$ converges

I am trying to prove that the series:

$$\sum^\infty_{n=1}\left( 1+ \frac{1}{n} \right)^n$$

converges.

Now I know that

$$\lim_{n\rightarrow\infty} \left( 1+ \frac{1}{n} \right)^n=e$$

But how can I use that knowledge to prove the convergance ?

Intuitively I would say that the series diverges since it doesn't approach zero but how can I formally prove this?

1. 2019-11-14

A necessary condition for the series $$\sum a_j$$ to converge is $a_j\to0$ as $j\to\infty$. If you can show that $a_j\not\to0$ as $j\to\infty$, then this implies that the series diverges. See here for more details.

2. 2019-11-14

$Proposition:$ If $\sum_{n=1}^{\infty}a_n$ converges then $a_n \rightarrow 0$

$Proof:$

Let $s_n= \sum_{k=1}^na_k$ then from convergence exists $s \in \mathbb{R}$ such that $s_n \rightarrow s$

Now we have that $a_n=s_n-s_{n-1} \rightarrow s-s=0$

3. 2019-11-14

hint

For $n>0$

$$\Bigl(1+\frac {1}{n}\Bigr)^n>1$$

and comparison test.

4. 2019-11-14

When $n$ is large, $(1+ 1/n)^n \approx e$ so the tail of your series looks like $$\cdots e + e + e + \cdots$$ so can't converge - as you suspected.