# Series Comparison test

• Sep 11th 2009, 09:06 PM
shinn
Series Comparison test
Hi, just wondering what can i compare: $\displaystyle \sum {\frac{1} {(ln (k))^9}}$ with to show it diverges?

Also, for $\displaystyle \sum { \frac{1} {k^{1 + \frac{1} {k}}} }$, how would i apprach this to test divergence/convergence?

• Sep 11th 2009, 11:33 PM
mr fantastic
Quote:

Originally Posted by shinn
[snip]
Also, for $\displaystyle \sum { \frac{1} {k^{1 + \frac{1} {k}}} }$, how would i apprach this to test divergence/convergence?

$\displaystyle k^{1 + \frac{1}{k}} = k \cdot k^{\frac{1}{k}}$. Now consider $\displaystyle \lim_{k \rightarrow \infty} k^{1 + \frac{1}{k}} = \lim_{k \rightarrow \infty} \left( k \cdot k^{\frac{1}{k}}\right) = \left( \lim_{k \rightarrow \infty} k\right) \cdot \left( \lim_{k \rightarrow \infty} k^{\frac{1}{k}} \right)$ and think about the necessary condition for the convergence of a series.
• Sep 12th 2009, 12:48 AM
shinn
Would this be correct?

Consider the inequality $\displaystyle \frac{1} {k^{1 + \frac{1}{k}}} < \frac{1}{k}$ as $\displaystyle 0 < \frac{1}{k} < 1$.

But by the p-series test, $\displaystyle \sum { \frac{1}{k}}$ diverges.
Hence, by the comparison test,$\displaystyle \sum { \frac{1} {k^{1 + \frac{1} {k}}} }$ must also diverge.
• Sep 12th 2009, 12:58 AM
mr fantastic
Quote:

Originally Posted by shinn
Would this be correct?

Consider the inequality $\displaystyle \frac{1} {k^{1 + \frac{1}{k}}} < \frac{1}{k}$ as $\displaystyle 0 < \frac{1}{k} < 1$.

But by the p-series test, $\displaystyle \sum { \frac{1}{k}}$ diverges.
Hence, by the comparison test,$\displaystyle \sum { \frac{1} {k^{1 + \frac{1} {k}}} }$ must also diverge.

To prove $\displaystyle \sum \frac{1} {k^{1 + \frac{1}{k}}}$ is divergent you require $\displaystyle \frac{1} {k^{1 + \frac{1}{k}}} {\color{red}>} f(k)$ where $\displaystyle \sum f(k)$ is divergent.

Re-examine my previous post.
• Sep 12th 2009, 02:35 AM
shinn
Hi,thanks for your help so far.
But, I don't really get the hint.
I know that a series diverges if you take the limit of the individual term to infinity and doesn't approach to zero.

But,
$\displaystyle \lim_{k \rightarrow \infty} \frac{1} {k^{1 + \frac{1}{k}}} = 0$
so its inconclusive?

Regarding my inequality, would it work if i place a fudge factor like:
$\displaystyle \frac{1}{2k}< \frac{1} {k^{1 + \frac{1}{k}}}$

Also adding to my first question:
What would be the "usual" comparison to compare something like $\displaystyle \sum {\frac{1} {(ln (k))^p}}$ , for different values of p? .
• Sep 12th 2009, 02:41 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
$\displaystyle k^{1 + \frac{1}{k}} = k \cdot k^{\frac{1}{k}}$. Now consider $\displaystyle \lim_{k \rightarrow \infty} k^{1 + \frac{1}{k}} = \lim_{k \rightarrow \infty} \left( k \cdot k^{\frac{1}{k}}\right) = \left( \lim_{k \rightarrow \infty} k\right) \cdot \left( \lim_{k \rightarrow \infty} k^{\frac{1}{k}} \right)$ and think about the necessary condition for the convergence of a series.

Quote:

Originally Posted by shinn
Hi,thanks for your help so far.
But, I don't really get the hint.
I know that a series diverges if you take the limit of the individual term to infinity and doesn't approach to zero.
But,

$\displaystyle \lim_{k \rightarrow \infty} \frac{1} {k^{1 + \frac{1}{k}}} = 0$

so its inconclusive?
[snip]

$\displaystyle \lim_{k \rightarrow \infty} k^{\frac{1}{k}} = 1$.

$\displaystyle \lim_{k \rightarrow \infty} k = + \infty$.

Now draw the obvious conclusion.
• Sep 12th 2009, 03:01 PM
TWiX
Sorry for this replay.
$\displaystyle \frac{1}{(\ln n)^9} > \frac{1}{n}$ for all n > m where m is a large enough number. So a proof using the comparison test is an obvious approach.