Question

Series help here.
Will make the questions on the next post.

Answer 1

I am having trouble proving that:
\[\sum_{n = 1}^{\infty} (-1)^{n-1}(\sqrt{n}/(n + 1))\]
converges.
I see it intuitively, but as I am trying to solve it using the Alternate series test, I am having trouble proving that bn+1 <= bn.

Answer 2

Similarly, I am having issues with:
\[\sum_{n = 1}^{\infty} (\sqrt{n + 1} - \sqrt{n - 1})/n\]

Answer 3

I got the intuition, but fail on prove it algebraically.

Answer 4

if you can show that

limit n-> inf tn = 0 and
tn+1 <= tn for some n => R, then the series converges.

I guess possibly it converges.

Answer 5

Indeed, but I am failing to show that tn+1 <= tn. My algebra manipulation is quite lame for these kind of problems.

Answer 6

I mean the limit of ratios ....

Answer 7

Also, for my second question, I tried to write it as:\[\sum_{n = 1}^{\infty} ((\sqrt{n + 1}/n) - (\sqrt{n-1}/n))\]
but fail to see how I can prove that sqrt(n+1)/n converges

Answer 8

Hmm, kay, will try it

Answer 9

from there it is possible to show such relation.

Answer 10

So, I get:\[\lim_{n \rightarrow \infty}|(\sqrt{n+1} (n+1))/((n+2)\sqrt{n})|\]right? I tried to simplify, but I am a bit stuck. I got\[\lim_{n \rightarrow \infty} |(n+1)^{3/2}/((n +2) \sqrt{n})|\]

Answer 11

I think we don't have to do this ....!!

Answer 12

I'll try to solve.

Answer 13

I thought that the way might be:\[\sqrt{n+1}/(n+2) \le \sqrt{n}/(n+1)\] then trying to square it. Don't know, tho

Answer 14

is,
\[\large \frac{2n^2 + 3n}{(n+1)^2} => 1 \] ???

Answer 15

Yes this is the way,

Answer 16

√n/(n+2)≤n−−√n+1/(n+1) square both sides, get my result,

show tat n^2+n >= 1 for all n => 1

Answer 17

that solves it.

Answer 18

Thanks mate :-)

Answer 19

yw