Series is convergent or divergent

**butbi9x** · March 15th 2009, 01:39 AM

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

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

**Moo** · March 15th 2009, 02:24 AM

Hello,

Originally Posted by butbi9x

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

$\lim_{n \to \infty} \frac{n+2}{n+1}=1 \neq 0$
hence this series diverges

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

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

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

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

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

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

Hence by comparison, the second series converges.

**butbi9x** · March 15th 2009, 03:10 AM

Can you Use the Integral Test to determine ?

**HallsofIvy** · March 15th 2009, 05:43 AM

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.

**Soroban** · March 15th 2009, 07:16 AM

Hello, butbi9x!

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

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

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

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

The series diverges.

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

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

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

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

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

The series converges.

Thread: Series is convergent or divergent

LinkBack

Thread Tools

Display

Series is convergent or divergent

Can you use Integral ?

Similar Math Help Forum Discussions

series convergent or divergent?

Show that the sum of a convergent and divergent series is divergent

Is this series convergent or divergent?

determine if series is absolutely convergent conditionally convergent or divergent

Series: Absolutely convergent, con convergent or divergent

Search Tags