Confusion about convergence tests There are two problems in particular I'm having trouble with.
1.) $$\sum_{n=1}^\infty(-1)^n\frac{\ln(n)}{\sqrt{n}}$$
I'm having difficulty demonstrating that $\frac{\ln(n+1)}{\sqrt{n+1}} \le \frac{\ln(n)}{\sqrt{n}}$.
Through simple cross multiplication I can get to $\ln(n+1)\sqrt{n} \le \ln(n)\sqrt{n+1}$, but I'm not sure how to prove that the inequality is true for all n after some value N.
2.) $$\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^{n^2}$$
I don't even know where to begin with this one. I tried using the Root Test as follows:
$$\lim_{n \to \infty}\sqrt[n^2]{\left(\frac{n}{n+1}\right)^{n^2}}$$
I'm actually not sure if I'm allowed to take the $n^2$ root of $|a_n|$ in the first place (some clarification on this would be much appreciated). Either way, using the root test this way just yielded a limit of 1, meaning the test is inconclusive. I don't know what else to try for this particular problem.
Also, two short questions about the validity of some algebraic manipulations I made in some problems I did earlier:
3.) Can I express $$\sum_{n=1}^\infty \frac{1*3*5*\cdot \cdot \cdot* (2n-1)}{2*5*8*\cdot \cdot \cdot * (3n-1)}$$ as $$\sum_{n=1}^\infty \frac{(2n-1)!}{(3n-1)!}$$
4.) In the same vein as above, can I express $$\sum_{k=1}^\infty \frac{(2^{k-1})(3^{k+1})}{k^k}$$
as
$$\sum_{k=1}^\infty \frac{(2^k)(3 * 3^k)}{2k^k}$$
Sorry if these seem obvious, but I've really been second guessing myself a lot since the textbook I'm working through began to cover series. The manipulations were, of course, made with the intent of facilitating the use of the ratio and root tests in 3) and 4) respectively.
 A: Define
$$f(x)=\frac{\log x}{\sqrt x}\implies f'(x)=\frac{\frac1{\sqrt x}-\frac{\log x}{2\sqrt x}}{x}=\frac{2-\log x}{2x\sqrt x}<0\iff \log x>2\iff x>e^2$$
and thus the function is monotonic descending for $\;x>e^2\;$ , and the sequence $\;\frac{\log n}{\sqrt n}\;$ is monotonic descending for $\;n>e^2\cong7.38\iff n\ge 8\;$
For two, with the $\;n-$ th root test:
$$\sqrt[n]{\frac1{\left(1+\frac1n\right)^{n^2}}}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<e$$
For (3). No, I don't think you can as you do not have succesive multiplying factors neither in the numerator nor in the denominator.
For(4). Yes, that is an algebraic identity, though I'd rather do
$$\sum_{k=1}^\infty\frac{2^{k-1}3^{k+1}}{k^k}=\frac32\sum_{k=1}^\infty\left(\frac6k\right)^k$$
and now you can use the root test, for example.
A: Hints:
1) Take the derivative of $\ln x/\sqrt x.$
2) The expression equals $[(1-1/(n+1))^{n+1}]^{n^2/(n+1)}.$
3) No, that quotient of factorials is not accurate. Instead, note that $(2k-1)/(3k-1) < 2/3, k = 1,2,\dots $
4) Yes, of course you can write it that way.
