# Thread: [SOLVED] Infinite Series Convergence

1. ## [SOLVED] Infinite Series Convergence

Hello, I have no idea how to show that the following infinite series diverge. I am supposed to compare then to the harmonic series... but I just don't get it.

1. $\sum ^\infty _{n=2} \frac{1}{(\log {n})^k}$

Hint $\frac {1}{(\log {n})^k} > \frac {1}{n}$ for a sufficiently large $n$

2. $\sum ^\infty _{n=1} \sin {\frac {1}{n} }$

Hint $\sin {\frac {1}{n}} > \frac {1}{2n}$ for a sufficiently large $n$

2. Originally Posted by akolman
Hello, I have no idea how to show that the following infinite series diverge. I am supposed to compare then to the harmonic series... but I just don't get it.

1. $\sum ^\infty _{n=2} \frac{1}{(\log {n})^k}$

Hint $\frac {1}{(\log {n})^k} > \frac {1}{n}$ for a sufficiently large $n$

2. $\sum ^\infty _{n=1} \sin {\frac {1}{n} }$

Hint $\sin {\frac {1}{n}} > \frac {1}{2n}$ for a sufficiently large $n$

For the first one look here

http://www.mathhelpforum.com/math-he...rgent-not.html

For the second one consider $\lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right )}{\frac{1}{n}}$

3. So is this correct?
$\lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right )}{\frac{1}{n}}
=\lim_{u\to 0}\frac{\sin (u)}{u}=1$
where $u=\frac{1}{n}$

4. Originally Posted by akolman
So is this correct?
$\lim_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right )}{\frac{1}{n}}
=\lim_{u\to 0}\frac{\sin (u)}{u}=1$
where $u=\frac{1}{n}$
Perfect....so we can assume that they share ____ by the ________ test