Determine whether the series converges or diverges.
\[\sum_{n = 1}^{\infty} (-1)^{n+1} \frac {\sqrt n + 1}{n+1}\]
\[\lim_{n \rightarrow \infty} a_n = 0\]But, how would I show that \[a_n > a_{n+1}\]If it is?
did you try finding the first derivative of the function f(x)=(sqrt(x)+1)/(x+1). If the derivative is negative than this function, and also the sequence which it represent is decreasing....with condition you wrote and this one ,you'll find that according to Leibnitz series conditionaly converges....
I already did the problem by showing that \[\frac {a_{k+1}}{a_k} < 1\] if n>0. But the derivative test works too I guess.
of course!!! good job...you can use any method....:D
If you'd like I can show you some formula which will help you very much if you're not sure if the solution is correct....of course, your teacher will always ask to prove, so you can use it JUST FOR CHECKING....
Oh, sure, that'd be great :)
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