# Check for the convergence or divergence of the following series

## Homework Statement

Here are some series I'm completely stuck on.
1.sqrt(n)*(1-cos(1/n))

2. a series in which if n is odd, then an is 1/(n+$\sqrt[]{n}$) while if n is even, then an is -1/n

## The Attempt at a Solution

For 1., I tried integral test which seemed impossible to integrate, and then I tried comparison test but i can't find anything to compare to

For 2. I thought that as n approaches infinity for 1/(n+$\sqrt[]{n}$), which equals (1/$\sqrt{n}$)/($\sqrt{n}$+1), then that expression pretty much looks like 1/n and thus the series is approx (-1)^n/n and is thus conditionally convergent like (-1)^n/n. But I then realized perhaps you can't apply the alternating series test since in the original series, the abs value of it is not decreasing for every term.E.g.1/(100+root(100))=1/110 < 1/101 Anyone have any idea on how to tackle this, or am I right from my original attempt?

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Zondrina
Homework Helper

## Homework Statement

Here are some series I'm completely stuck on.
1.sqrt(n)*(1-cos(1/n))

2. a series in which if n is odd, then an is 1/(n+$\sqrt[]{n}$) while if n is even, then an is -1/n

## The Attempt at a Solution

For 1., I tried integral test which seemed impossible to integrate, and then I tried comparison test but i can't find anything to compare to

For 2. I thought that as n approaches infinity for 1/(n+$\sqrt[]{n}$), which equals (1/$\sqrt{n}$)/($\sqrt{n}$+1), then that expression pretty much looks like 1/n and thus the series is approx (-1)^n/n and is thus conditionally convergent like (-1)^n/n. But I then realized perhaps you can't apply the alternating series test since in the original series, the abs value of it is not decreasing for every term.E.g.1/(100+root(100))=1/110 < 1/101 Anyone have any idea on how to tackle this, or am I right from my original attempt?

Your series is: ##\sum_{n=1}^{∞} \sqrt{n} - \sqrt{n}cos(\frac{1}{n})##

I know this may seem misleading, but this series appears to converge.

Hint: ##|cos(\frac{1}{n})| ≤ 1##

haruspex
Homework Helper
Gold Member
I know this may seem misleading, but this series appears to converge.

Hint: ##|cos(\frac{1}{n})| ≤ 1##
Yes, it converges, but you need to use a stronger fact about cos of small angles. Freshman2013, do you know the Taylor expansion of cos?

jbunniii
Homework Helper
Gold Member
For the first problem, this trig identity may be useful: ##2\sin^2(x) = 1 - \cos(2x)##.

jbunniii
Homework Helper
Gold Member
For problem 2, try adding ##a_n## and ##a_{n+1}##, where ##n## is odd.

so for the second one, could I do this:
the first term 1/(n+$\sqrt{n}$) for all odd n, I rewrite as a separate series 1/(2n-1+$\sqrt{2n-1}$) and the 2nd term -1/n for all even n I rewrite as -1/(2n)

I combine those two series and I get (1-$\sqrt{2n-1}$)/(4n^2-2n+2n$\sqrt{2n-1}$) and test for that(which is convergent). And I do the same except with 1/2n instead -1/2n to check for absolute convergence.

jbunniii