Converge? any aolternative methods of explaining?

I have this issue about some sums:

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{3^k-2^k}$

It must be divergent because $\displaystyle 3^k>2^k$ but I cant show that, any idéas?

Ohh yeah I tried with $\displaystyle 3^k=2^{k*ln(3)}$ but that didn´t really lead me anywhere,

And the second one is $\displaystyle \sum_{k=1}^{\infty}$$\displaystyle \frac{3k^2+3k+1}{k^3(k+1)^3}$

I get it to converge to 0 because the denominator goes towards infinity so much faster, but the books answer is 1. Dont know how to get that.

Any ideas what I am doing wrong?

Sorry but I dont understand

Quote:

Originally Posted by

**Krizalid** because i'd like to get an upper bound, so the numerator becomes bigger, then obviously for $\displaystyle k\ge1$ we have that $\displaystyle 3k^2+3k+1<3k^2+3k^2+k^2=7k^2.$

So you want an upperbound making the nominator larger, well in this case we ended upp with $\displaystyle \frac{7}{k^4}$

The books answer is $\displaystyle 1$.

How do I go from $\displaystyle \frac{7}{k^4}$ to $\displaystyle 1$?

Also my root test turned up like this, first time so no$\displaystyle \Rightarrow$ (Giggle)

$\displaystyle 3^k-2^k = 3^k(1-\frac{2^k}{3^k})$

$\displaystyle \lim_{x_\to \infty} [3^k(1-\frac{2^k}{3^k})]^{1/k}$

$\displaystyle

=\lim_{x_\to \infty} 3(1-\frac{2^k}{3^k})^{1/k}$

And here I´m stuck If I multiply by$\displaystyle \frac{1}{3^k}$ i get $\displaystyle [\frac{1}{3^k}-\frac{2^k}{3^{2k}}]$ so that didn´t work

But hey still have other algebraic idéas. Lets multiply by $\displaystyle \sqrt{3^k}$ ..and that gives me the original expression to the k:th root multiplied by 3...well....nja that doesnt help....(exept that I understand that if $\displaystyle (1-\frac{2^k}{3^k})^{1/k}<\frac{1}{3}$ then the limit is smaller then 1 and the series converges. Well great thats what I want to find out by still, im back where I started....lets try something else..and here I´m stuck....(Headbang)