# Thread: infinite sequences and series

1. ## 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!

2. 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

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

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

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

This is a geometric series. I hope this helps.

3. 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.

4. 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.

5. 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.