# Series Comparison test

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

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

Thanks in advance.
• September 11th 2009, 11:33 PM
mr fantastic
Quote:

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

Thanks in advance.

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

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

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

Originally Posted by shinn
Would this be correct?

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

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

Your argument is not correct.

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

Re-examine my previous post.
• September 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,
$\lim_{k \rightarrow \infty} \frac{1} {k^{1 + \frac{1}{k}}} = 0$
so its inconclusive?

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

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

Originally Posted by mr fantastic
$k^{1 + \frac{1}{k}} = k \cdot k^{\frac{1}{k}}$. Now consider $\lim_{k \rightarrow \infty} k^{1 + \frac{1}{k}} = \lim_{k \rightarrow \infty} \left( k \cdot k^{\frac{1}{k}}\right) = \left( \lim_{k \rightarrow \infty} k\right) \cdot \left( \lim_{k \rightarrow \infty} k^{\frac{1}{k}} \right)$ 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,

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

so its inconclusive?
[snip]

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

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

Now draw the obvious conclusion.
• September 12th 2009, 03:01 PM
TWiX
Sorry for this replay.
but please i want the answer of the first series ?
(1)/([ln n]^9)
• September 16th 2009, 04:45 AM
mr fantastic
Quote:

Originally Posted by TWiX
Sorry for this replay.
but please i want the answer of the first series ?
(1)/([ln n]^9)

$\frac{1}{(\ln n)^9} > \frac{1}{n}$ for all n > m where m is a large enough number. So a proof using the comparison test is an obvious approach.