# infinite sequences and series

• Oct 26th 2009, 06:10 PM
jsu03
infinite sequences and series
Can someone help me with these problems Thanks!

1. Determine whether the sequence converges or diverges, if converges, find limit. an = n/1+ squareroot n

I divided top and bottom by 1/n and got 1/0 + 1/0 = 0 = converges

2. find the sum of series e^n/3^n-1 I got no idea what to do, please help me, thanks again!
• Oct 26th 2009, 06:20 PM
TheEmptySet
Quote:

Originally Posted by jsu03
Can someone help me with these problems Thanks!

1. Determine whether the sequence converges or diverges, if converges, find limit. an = n/1+ squareroot n

I divided top and bottom by 1/n and got 1/0 + 1/0 = 0 = converges

2. find the sum of series e^n/3^n-1 I got no idea what to do, please help me, thanks again!

For the first one

$\frac{n}{1+\sqrt{n}}=\frac{1}{\frac{1}{n}+\frac{1} {\sqrt{n}}}$

Since the denominator goes to zero asn to infity the sequence divereges to infinity.

$\sum_{n=0}^{\infty}\frac{e^n}{3^{n-1}} =3\sum_{n=0}^{\infty}\frac{e^n}{3^{n}}$

This is a geometric series. I hope this helps.
• Oct 26th 2009, 06:21 PM
Debsta
For Q2, write out the first few terms of the sequence (ie let n=1, then n=2, etc) You'll see a pattern. What sort of sequence is it? Hint: what is a? r?. Then use the relevant formula to find the sum.
• Oct 26th 2009, 06:44 PM
jsu03
I got a= e/3, and r= 1/3 then, a/1-r = e/2 > 1 so the series diverges, is this even correct cause it look wrong to me.
• Oct 26th 2009, 07:04 PM
Debsta
a=e not e/3 using the formula you gave.
The sequence is (e^1)/1, (e^2)/3, (e^3)/9, (e^4)/27, ...
ie a = e, r = e/3
Then use a/(1-r) because r<1.