Question

Series Help: Prove the divergence of (1+sqrt(j))/(j(ln(j))^2) j=2 to inf

Answer 1

is that
\[ \sum_{j=2}^\infty \frac{1+\sqrt{j}}{j\cdot \ln^2(j)} \]
?

Answer 2

yes

Answer 3

This is pretty interesting, actually. First, notice the following... if you plot ln^2(x) and sqrt(x) on the same graph, you might see this: http://www.wolframalpha.com/input/?i=log^2%28j%29+%3D+j^.5

What that means is that there is some point beyond which sqrt(x) is greater than ln^2(x). It's quite large, somewhere around 5500 or so, but it exists, and that's all that matters. For all j beyond that point, we have that
\[ \ln^2(j) < \sqrt{j} \]
and therefore that
\[\frac{1+\sqrt{j}}{j \cdot \ln^2(j) } > \frac{1+\sqrt{j}}{j \cdot \sqrt{j}}\]
which is definitely greater than
\[ \frac{\sqrt{j}}{j\cdot \sqrt{j}} = \frac{1}{j} \]
So, for all j out way beyond that 5500 or so, we have that the terms in our series will be strictly greater than 1/j. But 1/j is the harmonic series, and it diverges, so by the comparison test (albeit a long, drawn out comparison) our series must also diverge.

Answer 4

Thank you so much!

Answer 5

Sure. Also, you could simply take
\[ \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{\ln^2(x)} = \lim_{x\rightarrow \infty}\frac{\frac{1}{2\sqrt{x}}}{\frac{2\ln(x)}{x}} = \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{4\ln(x)} = \lim_{x \rightarrow \infty} \frac{\frac{1}{2\sqrt{x}}}{\frac{4}{x}} = \lim_{x \rightarrow \infty} \frac{\sqrt{x}}{8} = \infty\]
via repeated application of L'Hospital's rule to show the same thing that I showed on the graph, which might be useful when doing this without a computer or graphing software.

Answer 6

Brilliant!