1. For which values of $p$ is the series $$\sum_{n=1}^\infty\frac1{(n (1+\ln n)^p)}$$ convergent/divergent?

    I've been trying to solve this earlier but the best I can come up with is the ratio test and I know that's not the proper way to solve for $p$.

  2. Where are each of these series convergent/divergent?

    1. $$\sum_{n=1}^\infty\frac1{n^a}-\frac1{(n+1)^a}$$
    2. $$\sum_{n=1}^\infty\ln\left(1+\frac1n\right)$$

    Again, I've been trying to do the ratio test but it's not working, mostly because of that extra variable that confuses me.



$1.$ Some useful facts:

If $p \gt 1$, then $\sum_1^\infty \frac{1}{n^p}$ converges. If this has not already been covered in your course, you can prove it using the Integral Test, or the Cauchy Condensation Test. But it is a very standard sort of series, often called a $p$-series.

$\sum_2^\infty \frac{1}{n\ln n}$ diverges. This is done for example using the Integral Test.

If I read your expression correctly, these facts and the Comparison Test should settle things.

By the way, $\sum_2^\infty \frac{1}{n(\ln n)^p}$ converges if $p\gt 1$, but you won't need this.

a) Find the sum of the first two terms. Find the sum of the first three. Find the sum of the first four. Whole lot of cancelling! Now you can write down a simple expression for the sum of the first $n$ terms. Under what conditions on $a$ does this converge to something finite?

b) Note that $\ln(1+1/n)=\ln((n+1)/n)=\ln(n+1)-\ln(n)$. Then use the same idea as in a).

  • $\begingroup$ You can get a list item with multiple paragraphs to be correctly formatted if you just indent each subsequent paragraph with a space. I can edit that into your answer, if you like. $\endgroup$ – Rahul Jan 27 '13 at 7:55
  • $\begingroup$ Thanks for the offer. The list only has two items, so maybe not worthwhile, but certainly fine with me. I am accustomed to TeX lists, and unfortunately this site only uses TeX for the mathematics. $\endgroup$ – André Nicolas Jan 27 '13 at 7:59
  • $\begingroup$ If the p-series says p>1 is always convergent isnt it the same for 2a?? it seems is should work $\endgroup$ – user1730308 Jan 28 '13 at 2:41
  • $\begingroup$ Certainly these work. But more do. If you followed my hint, you found that the sum of the first $n$ terms is $1-\frac{1}{(n+1)^a}$. If $a\gt 0$, this approaches $1$. And then there is the trivial $a=0$, where we are just adding up $0$'s. But if $a\lt 0$ then the term $\frac{1}{(n+1)^a}$ blows up. So the answer is $a\ge 0$ gives convergence, $a\lt 0$ gives divergence. $\endgroup$ – André Nicolas Jan 28 '13 at 2:55
  • $\begingroup$ Im guessing to find the first few sums you did telescoping? but i got 1/n^a - 1/(n+1)^a.... Am i doing it wrong? Also to get the sum what method would you suggest i use? $\endgroup$ – user1730308 Jan 28 '13 at 3:24

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