# Thread: What is the Limit and why?

1. ## What is the Limit and why?

Can somebody help me?

Find the lim of 1/n (1/2 + 2/3 + 3/4 + ... + n/n+1) as n tends to infinity.

Thanks!

2. hint:
$\frac12 + \frac23 +\frac34 .. +\frac{n}{n+1}$

=

$(1-\frac12) + (1-\frac13) + (1-\frac14).. +(1-\frac{1}{n+1})$

3. ## Ratio test?

You have the sum of n/(n+1) as n tends to infinty. You can use the ratio test on this if you know it, if not it is easy to observe that this is the same as the sum of 1/(1+1/n) as n tends to infinity (just divide top and bottom by n). So you can see as n tends to infinity the 1/n on the bottom tends to zero and you are left adding 1/1 or 1 onto the sum in each iteration of n. Therefore the sum diverges to infinity.

HTH,

Si

4. Originally Posted by bananaxxx
You have the sum of n/(n+1) as n tends to infinty. You can use the ratio test on this if you know it, if not it is easy to observe that this is the same as the sum of 1/(1+1/n) as n tends to infinity (just divide top and bottom by n). So you can see as n tends to infinity the 1/n on the bottom tends to zero and you are left adding 1/1 or 1 onto the sum in each iteration of n. Therefore the sum diverges to infinity.

HTH,

Si
You've considered only part of the expression. Clearly the part you've considered diverges because n(n + 1) --> 1 instead of zero (and so a necessary [but not sufficient] condition is not met).