# does the series \sum_{1}^{\infty} (\frac{n+1}{n})^n equals to e?

$$\sum_{1}^{\infty} (\frac{n+1}{n})^n$$

i used this solution to get the value of ab but it's wrong I think im performing the operations correctly :

$$ln(ab) = \lim_{n \to \infty} n *ln(\frac{n+1}{n})$$ $$ln(ab) = 1$$ $$ab = e$$

• I think you mean to refer to the sequence and not to the series. – lulu May 9 '17 at 11:32

The $n$-th term of the series does not go to $0$. We have that $$\biggl(\frac{n+1}n\biggr)^n=\biggl(1+\frac1n\biggr)^n\to e\ne0$$ as $n\to\infty$. Hence, the series does not converge.
• The divergence test : If the limit of $a_n$ is not zero, or does not exist, then the sum diverges; i.e., if $$\lim_{n \rightarrow \infty}a_n \neq 0.$$ then the series $\sum_{n=1}^{\infty} {a_n}$ diverges. – Ahmed May 9 '17 at 12:18