Converge?

**Henryt999** · February 11th 2010, 02:51 PM

I have this issue about some sums:

$\sum_{k=1}^{\infty} \frac{1}{3^k-2^k}$
It must be divergent because $3^k>2^k$ but I cant show that, any idéas?
Ohh yeah I tried with $3^k=2^{k*ln(3)}$ but that didn´t really lead me anywhere,

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

**Krizalid** · February 11th 2010, 02:57 PM

huh?

we can write $3^{k}-2^{k}=3^{k}\left( 1-\frac{2^{k}}{3^{k}} \right),$ now apply the root test, what do you get?

as for the second one, we can consider the following bounding: $\frac{3k^{2}+3k+1}{k^{3}(k+1)^{3}}<\frac{7k^{2}}{k ^{6}}=\frac{7}{k^{4}},$

**Henryt999** · February 11th 2010, 03:04 PM

Originally Posted by Krizalid

huh?

we can write $3^{k}-2^{k}=3^{k}\left( 1-\frac{2^{k}}{3^{k}} \right),$ now apply the root test, what do you get? Reading about it on wiki

as for the second one, we can consider the following bounding: $\frac{3k^{2}+3k+1}{k^{3}(k+1)^{3}}<\frac{7k^{2}}{k ^{6}}=\frac{7}{k^{4}},$

dont have a clue how you bounded and got $7k^2$ how?

**Krizalid** · February 11th 2010, 03:05 PM

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

**Henryt999** · February 11th 2010, 11:54 PM

Originally Posted by Krizalid

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

So you want an upperbound making the nominator larger, well in this case we ended upp with $\frac{7}{k^4}$
The books answer is $1$ .
How do I go from $\frac{7}{k^4}$ to $1$ ?

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

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

$\lim_{x_\to \infty} [3^k(1-\frac{2^k}{3^k})]^{1/k}$
$<br /> =\lim_{x_\to \infty} 3(1-\frac{2^k}{3^k})^{1/k}$
And here I´m stuck If I multiply by $\frac{1}{3^k}$ i get $[\frac{1}{3^k}-\frac{2^k}{3^{2k}}]$ so that didn´t work

But hey still have other algebraic idéas. Lets multiply by $\sqrt{3^k}$ ..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 $(1-\frac{2^k}{3^k})^{1/k}<\frac{1}{3}$ 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....

**JG89** · February 12th 2010, 10:09 AM

$3^k - 2^k = 3^k(1 - \frac{2^k}{3^k} ) > \frac{3^k}{2}$ for all sufficiently large $k$ .

Now compare your series to $\sum \frac{2}{3^k}$

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Converge? any aolternative methods of explaining?

well

Sorry but I dont understand

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