1. Convergence/Divergence of Series

Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.

Thanks for the help in advance

2. Originally Posted by jarny
Can someone give me a hint that the summation of (-1)^(u)*(u^(1/2))/(u+1) from u=1 to N is divergent, conditionally convergent, or absolutely convergent? By the looks of it it is conditionally convergent as it's absolute value diverges, but i know 1/u^(1/2)*(-1)^(u) converges and the one doesn't really matter, but I can't prove that it doesn't matter.

Thanks for the help in advance
You're right. It's conditionally convergent.

Your sum is $\sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+1}$. Considering $\sum_{i=1}^\infty \frac{\sqrt{n}}{n+1}$ show it's divergent by LCT with $\sum_{i=1}^\infty \frac{1}{\sqrt{n}}$.

The Alternating series test

(i) $\lim_{n \to \infty} \frac{\sqrt{n}}{n+1} = 0$
(ii) $\frac{\sqrt{n}}{n+1}$ is decreasing

give that it converges and thus, converges conditionally.

3. Hello, jarny!

$\sum^{\infty}_{u=1}\;(\text{-}1)^u\frac{u^{\frac{1}{2}}}{u+1}$ . Divergent, conditionally convergent, or absolutely convergent?
It is a convergent alternating series. .(The $n^{th}$ term approaches zero.)

It is not absolutely convergent because: . $\sum\frac{u^{\frac{1}{2}}} {u+1} \:\approx\:\sum\frac{1}{u^{\frac{1}{2}}}$ . . . a divergent $p$-series.

Therefore, the series is conditionally convergent.

4. Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.

when its (-1)^(u) *(u)^(1/2)/(u+4) could I just find a particular u (in this case 4) where the absolute value of the series begins to decrease and say since a_1>a_2>a_3 ...>a_N and since the lim n->infinity is zero (and the summands before 4 are finite) that by the alternating series test the summation is conditionally convergent?

5. Originally Posted by jarny
Somehow I managed to mis-write it. It in the denominator is should be 4 instead of 1. Sorry :/ Soroban, I can see that it is approximately but I need to kind of prove it.
If you mean

$
\sum_{i=1}^\infty \frac{(-1)^n \sqrt{n}}{n+4}$

It won't change the results that we have posted (except for the 4)
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