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

4 Answers

  1. Joseph- Reply

    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. Joshua- Reply

    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 $a_n=s_n-s_{n-1} \rightarrow s-s=0$

    Use this to derive a contradiction assuming that your series converges.

  3. Justin- Reply

    2019-11-14

    hint

    For $n>0$

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

    and comparison test.

  4. Keith- Reply

    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.

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