Hello, please help me show that this infinite sum converges (I

this is trivial.
Each term is decreasing and they are alternating in sign. It has to converge by the pendulum test.

>>8803174
Might I add, which is important, that the limit of each term is zero. If they were just decreasing monotonically, but had a lower bound, they wouldn't have to converge.

>>8803177
Isn't that the basic condition for convergence?

Yeah, it's trivial when you can do derivatives without mistakes (somehow the derivative was positive last time I calculated it).

>>8803268
>do derivates without mistakes
Okay, you're making a big mistake. When you're doing math not everything is about being rigorous or following the steps.
Just look at the fucking thing. It obviously is constantly decreasing. Use common sense.

>>8803270
to be fair, op did say "i think it does"

the thing is "common fucking sense, bro!" is not sufficient proof in most circles

>>8803138
evaluate
[math]
\lim_{n\to\infty}\frac{n+1}{n\log\left(n^2+1\right)} = 0
[/math]
so that by the alternating series test it converges

>>8803138
https://en.wikipedia.org/wiki/Alternating_series_test

>>8804049
Youre a retard. Having a limit be 0 doesnt mean it converges
>What is the harmonic series
Fucking kill yourself you goddamn brainlet

>>8804049
I believe the alternating series test will work.
>>8804372 is clearly a brainlet but they stumbled on a partially valid point. To use the alternating series test you need the terms to be eventually decreasing in absolute value.

These terms do decrease in absolute value, but you need to justify that.

>>8804372

Fucking disgusting brainlet,

>>8804049
quotes the alternating series theorem, the Leibniz criterium.

>>8804372
>>8804382
>>8804387
Nothing remotely wrong with what I said

>>8804406
There exist alternating series whose terms approach zero and diverge.

So there is something wrong with what you said.

The alternating series test I know requires that the terms also be decreasing in absolute value.

>>8803138

>tldr;
>help me pass my calc 2 final
>t. brainlet

>>8804372
>Having a limit be 0 doesnt mean it converges
i expressly invoked the ALTERNATING SERIES test to conclude that it converges you mongrel

>>8804499
then you better look up what you're invoking, because you fucked it up.

>>8805080
please explain

>>8805107
the terms have to be decreasing in absolute value have a limit of 0 and the sign has to alternate.

You only wrote the part about the alternating signs and the limit of 0 so there is still something to prove (even though it's simple)

>>8803177
If we have an alternating series

[math] \sum_{n=1}^\infty a_n [/math]

with

[math] a_n + (-1)\cdot a_{n+1} = \dfrac{1} {n} [/math]

then it wouldn't converge. How can we rule out that the above equation has a solution?

>>8805378
The theorem needs the absolute value of the general term [math]| a_n | [/math] to be decreasing, and that it goes to 0.

>>8805340
yeah, true, i figured it was clear enough that it is monotonically decreasing
thanks for keeping things real!

>>8805419
What's the name of the theorem?
Because it's not evident

[math] a_n + (-1)\cdot a_{n+1} = \dfrac{1} {n} [/math]

is rules out, just from [math] (a_n)_n [/math] being decreasing and going to zero.

https://en.wikipedia.org/wiki/Alternating_series_test

got it

>>8803138

Does the limit go to 0?
Are the positive terms non-increasing?

If yes to both, it converges.

Pay the Fuck attention in class. I hope to god youre not my tutee.