[SOLVED] Infinite Series Convergence

**akolman** · November 4th 2008, 06:29 PM

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 $n$

Thanks in advance.

**Mathstud28** · November 4th 2008, 06:37 PM

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 $n$

Thanks in advance.

For the first one look here

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

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

**akolman** · November 4th 2008, 06:45 PM

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

**Mathstud28** · November 4th 2008, 06:48 PM

Originally Posted by akolman

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

Perfect....so we can assume that they share ____ by the ________ test

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