# Thread: Series is convergent or divergent

1. ## Series is convergent or divergent

$\displaystyle \sum_{n=1}^{\infty }\frac{n+2}{n+1}$

$\displaystyle \sum_{n=1}^{\infty }ne^{-n}$

2. Hello,
Originally Posted by butbi9x
$\displaystyle \sum_{n=1}^{\infty }\frac{n+2}{n+1}$
$\displaystyle \lim_{n \to \infty} \frac{n+2}{n+1}=1 \neq 0$
hence this series diverges

$\displaystyle \sum_{n=1}^{\infty }ne^{-n}$
$\displaystyle e^n=\underbrace{1+n+\frac{n^2}{2}}_{\geq 0}+\frac{n^3}{6}+\underbrace{\frac{n^4}{24}+\dots} _{\geq 0}$

Hence $\displaystyle e^n \geq \frac{n^3}{6}$

---> $\displaystyle e^{-n}=\frac{1}{e^n} \leq \frac{6}{n^3}$

---> $\displaystyle ne^{-n} \leq \frac{6}{n^2}$

and $\displaystyle \sum_{n=1}^\infty \frac{6}{n^2}$ converges (Riemann series).

Hence by comparison, the second series converges.

3. ## Can you use Integral ?

Can you Use the Integral Test to determine ?

4. Yes, it would be a very good method to use. In fact, because you thought of it, it would be better for you to use it rather than Moo's method.

The difficulty is that since you chose not to show what you had done or tried when you posted the problem, we cannot tell what response is best for you.

5. Hello, butbi9x!

Use Integral Test to determine whether the series is connvergent or divergent:

. . . $\displaystyle \sum_{n=1}^{\infty }\frac{n+2}{n+1}$

We use: .$\displaystyle \int^{\infty}_1\frac{x+2}{x+1}\,dx \;=\;\int^{\infty}_1\left(1 + \frac{1}{x+1}\right)\,dx \;=\;x + \ln(x+1)\,\bigg]^{\infty}_1$

We have: .$\displaystyle \lim_{b\to\infty}\bigg[x + \ln(x+1)\bigg]^b_1 \;=\;\lim_{b\to\infty}\bigg[b + \ln(b+1)\bigg] \;=\;\infty$

The series diverges.

$\displaystyle \sum_{n=1}^{\infty}ne^{-n}$

We use: .$\displaystyle \int^{\infty}_1 xe^{-x}\,dx$

Integrate by parts: .$\displaystyle \begin{array}{ccccccc}u &=& x & & dv &=& e^{-x}\,dx \\ du &=& dx & & v &=&-e^{-x} \end{array}$

We have: .$\displaystyle -xe^{-x} + \int e^{-x}\,dx \;=\;-xe^{-x} - e^{-x}\,\bigg]^{\infty}_1 \;=\;-e^{-x}(x+1)\,\bigg]^{\infty}_1$

Then: .$\displaystyle \lim_{b\to\infty} \bigg[-\frac{x+1}{e^x}\,\bigg]^b_1 \;=\;\lim_{b\to\infty}\bigg[-\frac{b+1}{e^b} + \frac{2}{e}\bigg] \;=\;-0 + \frac{2}{e} \;=\;\frac{2}{e}$

The series converges.