- #1

DryRun

Gold Member

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## Homework Statement

[tex](a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}\;(solved)[/tex]

[tex](b)\;\sum^{\infty}_{r=1}\frac{2r}{1+r^2}\;(solved)[/tex]

[tex](c)\;\sum^{\infty}_{n=1}\frac{\cos^4 nx}{n^2}\;(solved)[/tex]

[tex](d)\;\sum^{\infty}_{n=1}\frac{3^r+4^r}{4^r+5^r}[/tex]

[tex](e)\;\sum^{\infty}_{r=1}\frac{r^r}{r!}\;(solved)[/tex]

[tex](f)\;\sum^{\infty}_{n=1}\frac{1.2.3...n}{4.7.10... (3n+1)}\;(solved)[/tex]

## Homework Equations

nth-term test, ratio test, comparison test, limit comparison test, geometric series, etc.

## The Attempt at a Solution

[tex](a)\;\sum^{\infty}_{n=1}\frac{n-5}{n^2}=\sum^{\infty}_{n=1}\frac{1}{n}-\sum^{\infty}_{n=1}\frac{5}{n^2}[/tex]This is contradictory, as according to the p-series test, the first series diverges but the second series converges. I also tried to factorize out ##n## or ##n^2## out of the numerator and denominator, but got limit = 0, which is inconclusive.

[tex](b)\;\lim_{r\to \infty}\frac{2}{1/r+r}=0[/tex]I factorized ##r## out of the numerator and denominator, then took the limit but got 0, which is inconclusive. I also tried comparison test, since ##v_n=2/n \ge u_n## and the series ##v_n## diverges, the test is inconclusive.

[tex](c)\;0\le \frac{\cos^4 nx}{n^2}\le 1/n^2[/tex]

I used the squeeze theorem on the sequence, but got 0, which is inconclusive.

[tex](d)\;\lim_{n\to \infty}\frac{1+(3/4)^r}{1+(5/4)^r}=1/∞=0[/tex]

Since the limit = 0, it is inconclusive.

[tex](e)\;\lim_{r\to \infty}\left( \frac{r+1}{r} \right)^r[/tex]

I used the ratio test. Then got stuck. Took logs and got the limit = ##e^∞##?

[tex](f)\;\lim_{n\to \infty}\frac{(4+4/n)}{(3+4/n)}=4/3[/tex]

I used the ratio test. L>1, so the series diverges. But the answer is convergent series.

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