[SOLVED] Two Series for Convergence

**Paperwings** · August 18th 2008, 05:44 AM

Problem
Examine the following series for convergence
1. $\sum_{k=1}^{\infty}ke^{-k^2}$
2. $\sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3$
=================================
Attempt:
1. This series $\sum_{k=1}^{\infty}ke^{-k^2}$ can be rewritten as $\sum_{k=1}^{\infty} \frac{k}{e^{k^2}}$ .
I try to use the Ratio Test. By the Ratio Test

$\lim_{k \to \infty} \left| \frac{k+1}{e^{(k+1)^2}} \frac{e^{k^2}}{k}\right| = \lim_{k \to \infty} \left| \frac{e^{k^2}}{e^{(k+1)^2}} \left( 1 + \frac{1}{k}\right) \right|$

I know the term 1 + 1/k goes to 1 as $k \rightarrow \infty$ . However I do know about the left term. $\frac{e^{k^2}}{e^{(k+1)^2}}$

2. I tried to find if this series converges or not by the Comparison Test but went nowhere.

Thank you for your time.

**kalagota** · August 18th 2008, 05:50 AM

Originally Posted by Paperwings

Problem
Examine the following series for convergence
1. $\sum_{k=1}^{\infty}ke^{-k^2}$
2. $\sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3$
=================================
Attempt:
1. This series $\sum_{k=1}^{\infty}ke^{-k^2}$ can be rewritten as $\sum_{k=1}^{\infty} \frac{k}{e^{k^2}}$ .
I try to use the Ratio Test. By the Ratio Test

$\lim_{k \to \infty} \left| \frac{k+1}{e^{(k+1)^2}} \frac{e^{k^2}}{k}\right| = \lim_{k \to \infty} \left| \frac{e^{k^2}}{e^{(k+1)^2}} \left( 1 + \frac{1}{k}\right) \right|$

I know the term 1 + 1/k goes to 1 as $k \rightarrow \infty$ . However I do know about the left term. $\frac{e^{k^2}}{e^{(k+1)^2}}$

2. I tried to find if this series converges or not by the Comparison Test but went nowhere.

Thank you for your time.

$\frac{e^{k^2}}{e^{(k+1)^2}} = \frac{e^{k^2}}{e^{k^2 + 2k + 1}} = \frac{e^{k^2}}{e^{k^2}e^{2k}e} \longrightarrow ...$

**kalagota** · August 18th 2008, 05:54 AM

Originally Posted by Paperwings

2. $\sum_{k=1}^{\infty} \left(\frac{k+1}{k^2+1}\right)^3$
=================================

$\left(\frac{k+1}{k^2+1}\right)^3 < \left(\frac{k+1}{k^2}\right)^3 = \frac{k^3 + 3k^2 + 3k + 1}{k^6} ...$

**Paperwings** · August 18th 2008, 06:31 AM

Originally Posted by kalagota

$\frac{e^{k^2}}{e^{(k+1)^2}} = \frac{e^{k^2}}{e^{k^2 + 2k + 1}} = \frac{e^{k^2}}{e^{k^2}e^{2k}e} \longrightarrow ...$

Then $\frac{e^{k^2}}{e^{k^2}e^{2k}e} = \frac{1}{e^{2k+1}}\rightarrow 0$
Since 0 <1, then the series converges absolutely by the Ratio Test.

For the second one, from your hint
$\frac{k^3 + 3k^2 + 3k + 1}{k^6} = \frac{1}{k^3}+\frac{3}{k^4}+\frac{3}{k^5}+\frac{1} {k^6}$ . Then by the p-Test ( the series $\sum_{k=1}^{\infty} \frac{1}{k^p}$ converges if and only if p > 1) since all k's are > 1 then the series converge.

Thank you Kalagota.

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