To converge or to diverge, thats the q

[mynameisfunk]

Prove or find a counterexample:
(i) If \sum a_n converges, then \sum \frac{a_n}{\sqrt{n}} converges.
(ii) If \sum a_n converges, then \sum \frac{|a_n|}{n} converges.

for (i) I really haven't much of a clue.
for (ii) I am also confused but my thinking was that I could take a series \sum a_n that is not absolutely convergent and then since two divergent series are multplied together then they must diverge. However, I know that if I take a_n=\frac{(-1)^n}{n}that this won't work out since \sum \frac{1}{n^2} converges...

I would really like some help on this, and if you could i would like to be briefed on what a good strategy is for thinking about these problems, it seems like the amount of tests and manipulations that are possible are very overwhelming.

 

[JG89]

This may be overkill, but... theorem:

If \sum a_n converges and if b_1, b_2, ... is a bounded monotonic sequence of numbers, then \sum a_n b_n converges.

 

[fourier jr]

or the comparison test

 

[lgd0612]

Number one, using comparison test to prove that the sequence converges.

Number two, construct a monotone decreasing sequence out of the given sequence such that the new sequence is less or equal to the given sequence, then use the theorem below, to show that the constructed series diverges, thus the given sequence diverges

(this is my first post, and i don't know why my latex isn't working right, sorry. )

 

[lgd0612]

OK this is a test,
ill try to rewrite the theorem in latex

For any given monotone decreasing sequence {a_n}, the series \sum a_n converges
if and only if
\sum 2^k a_{2^k}, k=0,1,2,3,... converges.

\sum 2^k a_{2^k} = a_1+2a_2+4a_4+8a_8+...

Remark: this theorem works great when you have n (the index) in the denominator, because it cancels out with the 2^k in the front!

for example: the series \sum \frac{b_n}{n} converges if and only if

\sum 2^k \frac{b_{2^k}}{2^k} = 1\frac{b_1}{1}+2\frac{b_2}{2}+4\frac{b_4}{4}...= b_1+b_2+b_4+b_8+... converges.

you can try to prove \sum 1/n diverges using this theorem as a practice to get familiar with it.