Question

hi can someone help me see if these 2 series are convergent or divergent
1)
2^n
------
(n^n)-1
and
2)
1
----
(1+1/2)(1+1/3)...(1+1/n)

Answer 1

\[\sum_{n=2}^{\inf} 2^n/(n^n -1 ) \]

Answer 2

\[\sum_{n=1}^{\inf} 1\div((1+1/2)(1+1/3)...(1+1/n)) \]

Answer 3

For the first series, you can try either the ratio or root test for convergence. The root test is more easy on the eyes, I think:
\[\lim_{n\to\infty}\sqrt[n]{\left|\frac{2^n}{n^n-1}\right|}\]
If this limit is less than 1, then the series is convergent.

Alternatively, if you try the ratio test:
\[\lim_{n\to\infty}\left|\frac{2^{n+1}}{(n+1)^{n+1}-1}\cdot\frac{n^n-1}{2^{n}}\right|\]
Similarly, if the limit is less than 1, the series converges.

Answer 4

That makes sense

Answer 5

but for the first equation after you apply root n do you get \[2\div(n-1) ?\]

Answer 6

Because i did it that way but i wasn't sure what root n equaled

Answer 7

Not exactly, you have to apply certain reasoning to simplify the limit computation:
\[\lim_{n\to\infty}\left|\frac{2^n}{n^n-1}\right|^{1/n}=\lim_{n\to\infty}\left|\frac{2^n}{n^n}\right|^{1/n}\]
This is true because the \(-1\) in the denominator doesn't have that great an effect on the value of the denominator. Both \(n^n-1\) and \(n^n\) approach infinity at the same rate. Next,
\[\lim_{n\to\infty}\left|\frac{2^n}{n^n}\right|^{1/n}=\lim_{n\to\infty}\left|\frac{2}{n}\right|^{n/n}=\lim_{n\to\infty}\frac{2}{n}\]
Here you just apply some exponent properties. Obviously the limit is 0, which is less than 1, so the series converges.

Answer 8

Ohh so it's pretty simple

Answer 9

But for the second equation is the answer convergent because it's 1/(1+1/n))!? leading to
1/infinity ? Leasing to 0?

Answer 10

Hmm, I don't follow your reasnoing. The given series is not the same as \(\displaystyle\sum\frac{1}{1+\frac{1}{n}}\). Consider the first few terms of this series and the given one:
\[\begin{align*}\sum_{n=1}^{\infty}\frac{1}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{n}\right)}&=\frac{1}{\left(1+\frac{1}{2}\right)}\\&~~~~+\frac{1}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)}\\&~~~~+\frac{1}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)}+\cdots\\
&=\frac{1}{\frac{3}{2}}+\frac{1}{\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)}+\frac{1}{\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\left(\frac{5}{4}\right)}+\cdots\\
&=\frac{2}{3}+\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)+\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)+\cdots
\end{align*}\]
\[\begin{align*}\sum_{n=1}^{\infty}\frac{1}{1+\frac{1}{n}}&=\frac{1}{1+1}+\frac{1}{1+\frac{1}{2}}+\frac{1}{1+\frac{1}{3}}+\cdots\\
&=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots\end{align*}\]
You should be able to see why they're not equal.

Here's what I think you should do. Rewrite the series:
\[\begin{align*}\sum_{n=1}^{\infty}\frac{1}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdots\left(1+\dfrac{1}{n}\right)}&=\sum_{n=1}^{\infty}\frac{1}{\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)\cdots\left(\dfrac{n+1}{n}\right)}\\
&=\sum_{n=1}^{\infty}\left(\dfrac{2}{3}\right)\left(\dfrac{3}{4}\right)\cdots\left(\dfrac{n}{n+1}\right)\end{align*}\]
Now, apply the limit test for divergence; that is, given \(\sum a_n\), if \(\displaystyle\lim_{n\to\infty}a_n\not=0\), then the series diverges.

Answer 11

So in this case factorials aren't used at all ?

Answer 12

You could write it as a factorial product, sure:
\[\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\cdots\left(\frac{n}{n+1}\right)=2\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\cdots\left(\frac{n}{n+1}\right)=2\frac{n!}{(n+1)!}\]
Now you could easily apply the ratio test, if you'd like. You'll get the same result with either test.

Answer 13

Ok i see where i went wrong in my reasoning

Answer 14

Thanks a lot !! This was extremely helpful

Answer 15

You're welcome