1. ## Infinite Series.

Classify the following series as convergent or divergent. Provide justification (eg. which convergence test used.). For those series containing symbol $\alpha$ explain clearly for which real values of $\alpha$ if any the series converges or diverges.

(a) $\sum^\infty_{n=1} \frac{1}{3^n}{( 1 + \frac{1}{n})}^n$
(b) $\sum^\infty_{n=1} \frac{(-1)^{n-1}}{n^{1 + \alpha}}$
(c) $\sum^\infty_{n=0} \frac{(-1)^n n!}{(n+1)^{\alpha}}$
(d) $\sum^\infty_{n=2} \frac{1}{(n log n)^{\alpha}}$

And please kindly check if my working below is accurate or if I am on the right track.

The question is to evaluate the improper integral $\int^\infty_0 x^3 exp(-x^2) dx$.

Workings :

$\int^t_0 x^3e^{-x^2} dx$

Integration by Substitution,
Let $w = x^2$

$dw = 2x dx$

$dx = \frac{dw}{2x}$

At $x = t, w = t^2$

$x = 0, w = 0$

$\Rightarrow \int^{t^2}_0 w.xe^{-w} \frac{dw}{2x}
= \frac{1}{2} \int^{t^2}_0 we^{-w} dw$

Integration by Parts,
Let $u = w, \frac{du}{dw} = 1$
$\frac{dv}{dw} = e^{-w}, v = -e^{-w}$

$\Rightarrow \frac{1}{2}[ -t^2e^{-t^2} - (e^{-t^2} - 1)]$

Allowing $t \rightarrow \infty$
$\Rightarrow \frac{1}{2} [0 + 1] \rightarrow \frac{1}{2}$.

Thank you!

2. I can't remember how to do series questions but the integral you state and your working looks okay to me.

3. Hello,

For a), use the root test
For b), use the alternating series or compare it to the Riemann alternating series ( $\sum \frac{(-1)^n}{n^\alpha}$ converges iff $\alpha \geqslant 1$)
For c), use the ratio test
For d), you can look here for Bertrand series

4. Originally Posted by Moo
Hello,

For a), use the root test
For b), use the alternating series or compare it to the Riemann alternating series ( $\sum \frac{(-1)^n}{n^\alpha}$ converges iff $\alpha \geqslant 1$)
For c), use the ratio test
For d), you can look here for Bertrand series
Hello, I don't really understand the root test as I didn't learn that in my lectures.

As for part (c),
I got

$\lim_{n\rightarrow\infty} |\frac{(-1)(n+1)^{\alpha+1}}{(n+2)^\alpha}|$

Am I doing the correct thing, if so, how do I proceed from here?

I have not learnt the Bertrand series before either.

The list of test that I learnt was the divergence (DWMT) test, Comparison test, Limit Comparison test, Alternating Series test and ratio test.

From the above tests I learned, which is applicable? I'll keep trying and try to learn up the tests you suggested, but hopefully I'd be able to apply the tests I learned!

Hope to hear from you soon. Thank you!

5. Originally Posted by pearlyc
Hello, I don't really understand the root test as I didn't learn that in my lectures.
Hmm it's just an equivalent of the ratio test :
if $\lim_{n \to \infty} ~ |a_n|^{1/n} < 1$ then it converges (I admit it is not the first formula the wikipedia gives)

for a), I have another way to propose to you.
You have $a_n=\left(\frac{1+\frac 1n}{3}\right)^n$
Noting that $n \geqslant 1$, you can say that $1+\frac 1n \leqslant 2$ for all n.

So $a_n \leqslant \left(\frac 23\right)^n$
Does this series converge ?

As for part (c),
I got

$\lim_{n\rightarrow\infty} |\frac{(-1)(n+1)^{\alpha+1}}{(n+2)^\alpha}|$

Am I doing the correct thing, if so, how do I proceed from here?
Ooops, I'm sorry ! Actually, I guess it's better to use an alternating series test (I missed the (-1)^n part )

I have not learnt the Bertrand series before either.
In fact, I put this link so that you can see how they deal with such functions, how they conclude why it converges or diverges. It's similar here.

The list of test that I learnt was the divergence (DWMT) test, Comparison test, Limit Comparison test, Alternating Series test and ratio test.

From the above tests I learned, which is applicable? I'll keep trying and try to learn up the tests you suggested, but hopefully I'd be able to apply the tests I learned!

Hope to hear from you soon. Thank you!
Excuse me in advance, but I don't know how to use the integral test, so maybe you can use it for one or two series here. Feel free to try

6. Originally Posted by Moo
Hmm it's just an equivalent of the ratio test :
if $\lim_{n \to \infty} ~ |a_n|^{1/n} < 1$ then it converges (I admit it is not the first formula the wikipedia gives)

for a), I have another way to propose to you.
You have $a_n=\left(\frac{1+\frac 1n}{3}\right)^n$
Noting that $n \geqslant 1$, you can say that $1+\frac 1n \leqslant 2$ for all n.

So $a_n \leqslant \left(\frac 23\right)^n$
Does this series converge ?

Ooops, I'm sorry ! Actually, I guess it's better to use an alternating series test (I missed the (-1)^n part )

In fact, I put this link so that you can see how they deal with such functions, how they conclude why it converges or diverges. It's similar here.

Excuse me in advance, but I don't know how to use the integral test, so maybe you can use it for one or two series here. Feel free to try
For the alternating test for (b) is the $b_n = \frac{1}{n^{1+\alpha}}$ and we prove that this is convergent/divergent hence conclude by alternating test?

For the (c), I reckon the $b_n = \frac{n!}{(n+1)^{\alpha}}$ right and this should be divergent because factorials grow at a much faster rate hence by alternating test this series is divergent?

Hmm I am not too sure actually, and what do they mean by explain clearly for which real values of $\alpha$ if any the series converges and for which real values of $\alpha$ the series diverges?