## Review: Convergence of Series

Test the following series for convergence or divergence:

 $\displaystyle{\sum_{n=1}^\infty \frac{1}{n+3^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{(2n+1)^n}{n^{2n}}}$ $\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{n}{n+2}}$ $\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{n}{n^2+2}}$ $\displaystyle{\sum_{n=1}^\infty \frac{n^2 2^{n-1}}{(-5)^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{1}{2n+1}}$ $\displaystyle{\sum_{n=2}^\infty \frac{1}{n \sqrt{\ln n}}}$ $\displaystyle{\sum_{n=1}^\infty \frac{2^n n!}{(n+2)!}}$ $\displaystyle{\sum_{n=1}^\infty n^2 e^{-n}}$ $\displaystyle{\sum_{n=1}^\infty n^2 e^{-n^3}}$ $\displaystyle{\sum_{n=2}^\infty \frac{(-1)^{n+1}}{n \ln n}}$ $\displaystyle{\sum_{n=1}^\infty \sin n}$ $\displaystyle{\sum_{n=1}^\infty \frac{3^n n^2}{n!}}$ $\displaystyle{\sum_{n=1}^\infty \frac{\sin 2n}{1+2^n}}$ $\displaystyle{\sum_{n=0}^\infty \frac{n!}{2 \cdot 5 \cdot 8 \cdot \dotsb \cdot (3n+2)}}$ $\displaystyle{\sum_{n=1}^\infty \frac{n^2+1}{n^3+1}}$ $\displaystyle{\sum_{n=1}^\infty (-1)^n 2^{1/n}}$ $\displaystyle{\sum_{n=2}^\infty \frac{(-1)^{n-1}}{\sqrt{n} - 1}}$ $\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{\ln n}{\sqrt{n}}}$ $\displaystyle{\sum_{n=1}^\infty \frac{n+5}{5^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{(-2)^{2n}}{n^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{\sqrt{n^2-1}}{n^3+2n^2+5}}$ $\displaystyle{\sum_{n=1}^\infty \tan (1/n)}$ $\displaystyle{\sum_{n=1}^\infty n \sin(1/n)}$ $\displaystyle{\sum_{n=1}^\infty \frac{n!}{e^{n^2}}}$ $\displaystyle{\sum_{n=1}^\infty \frac{n^2+1}{5^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{n \ln n}{(n+1)^3}}$ $\displaystyle{\sum_{n=1}^\infty \frac{e^{1/n}}{n^2}}$ $\displaystyle{\sum_{n=1}^\infty \frac{(-1)^n}{\cosh n}}$ $\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n+5}}$ $\displaystyle{\sum_{n=1}^\infty \frac{5^n}{3^n+4^n}}$ $\displaystyle{\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}}$ $\displaystyle{\sum_{n=1}^\infty \frac{\sin(1/n)}{\sqrt{n}}}$ $\displaystyle{\sum_{n=1}^\infty \frac{1}{n+n\cos^2 n}}$ $\displaystyle{\sum_{n=1}^\infty \bigg( \frac{n}{n+1} \bigg)^{n^2}}$ $\displaystyle{\sum_{n=2}^\infty \frac{1}{\big( \ln n \big)^{\ln n}}}$ $\displaystyle{\sum_{n=1}^\infty \big( 2^{1/n} -1 \big)^n}$ $\displaystyle{\sum_{n=1}^\infty \big( 2^{1/n} -1 \big)}$
1. November 3, 2011 at 2:40 pm

On number five, I did by the ratio test. However, for a hypothetical example similar, let’s say the ratio test gives a limit of less than one, but the alternating series test shows the series increasing. Which test do I listen to, which takes precedence? Or is this even possible?

• November 3, 2011 at 8:50 pm

That is an excellent question! Alternating series test takes precedence over ratio/root.

2. November 5, 2011 at 3:11 pm

for the alternating series test, if both conditions are null, then does that automatically mean that the series is divergent, OR simply that it is not convergent and the test for diveregence needs to be done?

• November 5, 2011 at 7:58 pm

As soon as one of the conditions is not satisfied, the test cannot be applied. You need to use ratio or root instead.

3. November 5, 2011 at 3:59 pm

also for alternating series problems do we need to specify whther they or conditionally or absolutely convergent

• November 5, 2011 at 7:58 pm

yes

4. November 7, 2011 at 8:20 pm

For number 21, you do the root test, but does the series also have to satisfy the alternating series test?

• November 7, 2011 at 8:42 pm

21 is not an alternating series.

• November 7, 2011 at 8:45 pm

Whoops, I see that now. But if it was and it was decreasing, you could do the ratio test correct?

• November 8, 2011 at 6:56 am

If the limit of the sequence is zero too, yes.

1. No trackbacks yet.