# Thread: [SOLVED] does the series converge

1. ## [SOLVED] does the series converge

In my calc 3 class were using the divergence and integral test to determine wheter series converge or diverge. Im drawing blanks on this problem !

K / ln(k+1)

2. Originally Posted by nikie1o2
In my calc 3 class were using the divergence and integral test to determine wheter series converge or diverge. Im drawing blanks on this problem !

K / ln(k+1)
If

$\displaystyle \lim_{x \to \infty} \frac{x}{\ln(x+1)} \ne 0$ the series diverges. Use L'Hopital's rule to show this doesn't happen, i.e.

$\displaystyle \lim_{x \to \infty} \frac{x}{\ln (x+1)} \to \infty$

3. Hello, nikie1o2!

Converge or diverge? . $\displaystyle S \;=\;\sum^{\infty}_{k=1} \frac{k}{\ln(k+1)}$
I'd use the Comparison Test . . .

We have: .$\displaystyle \ln(k+1) \:<\:k+1$

Then: .$\displaystyle \frac{\ln(k+1)}{k} \:<\:\frac{k+1}{k}\quad \text{ for positive }k$

Invert: .$\displaystyle \frac{k}{\ln(k+1)} \:>\:\frac{k}{k+1}$

Take sums: .$\displaystyle S \;=\;\sum^{\infty}_{k=1}\frac{k}{\ln(k+1)} \;> \;\sum^{\infty}_{k=1}\frac{k}{k+1}$

Now: .$\displaystyle \sum^{\infty}_{k=1}\frac{k}{k+1} \;>\;\sum^{\infty}_{k=1}\frac{1}{k+1} \;=\;\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots$ divergent Harmonic Series

Hence: .$\displaystyle S \:>\:\text{(divergent series)}$

Therefore: .$\displaystyle S$ diverges.