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Let $displaystyle sum a_n$ be convergent series of positive terms and set $displaystyle b_n=sum_k=n^inftya_n$ , then prove that $displaystylesum fraca_nb_n$ diverges.

I could see that $b_n$ is monotonically decreasing sequence converging to $0$ and I can write $displaystylesum fraca_nb_n=sumfracb_n-b_n+1b_n$, how shall I proceed further?

Sorry, my previous answer was not correct. A new tentative:

$$fracb_k+1b_k=1-fraca_kb_k$$
Hence
$$fracb_N+1b_1=prod_k=1^N(1-fraca_kb_k)$$ and
$$log b_N+1-log b_1=sum_k=1^N log(1-fraca_kb_k)$$
Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k to 0$, and as $log(1-x)sim -x$ and the series have constant sign, this imply that the series $displaystyle log (1-fraca_kb_k)$ is convergent, a contradiction as $log (b_N+1) to -infty$.

Assuming $sum a_n=L$, for any $n$ big enough we must have:
$$a_n geq frac1nsum_m>na_m,tag1$$
otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have:
$$L< a_N+frac1N(L-a_n)+fracN-1(N+1)N(L-a_n)+ldots = L,tag2$$
contradiction. This implies that for any $ngeq M$
$$left(1+frac1nright)a_ngeqfrac1nb_ntag3$$
holds, hence:
$$sum_ngeq Mfraca_nb_ngeqsum_ngeq Mfrac1n+1,tag4$$
but the RHS of $(4)$ diverges.

First note that $b_n_ninmathbb N$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$
$$
fraca_mb_m+cdots+fraca_nb_nge
fraca_mb_m+cdots+fraca_nb_m=frac1b_m(a_m+cdots+a_n)=fracb_n-b_mb_m=1-fracb_nb_m.
$$
Next, as $b_nsearrow 0$, choose $m_1,m_2,ldots,m_k,ldots$, so that
$$
fracb_m_i+1b_m_i<1/2.
$$
Then we have that
$$
sum_n=1^m_kfraca_nb_ngesum_i=1^k-1sum_n=m_i+1^m_i+1fraca_nb_nge
sum_i=1^k-1left(1-fracb_m_ib_m_i+1right)gefrack-12,
$$
and hence $displaystylesum_n=1^inftyfraca_nb_n=infty$.

For $m>n$ one has
$$beginalignedfraca_nb_n+cdotsfraca_mb_m&ge fraca_nb_n+cdots +fraca_mb_n\
&=fracb_n-b_m+1b_n = 1-fracb_m+1b_n.
endaligned$$
Can you continue from here?

For a sum $sum_k=0^inftyc_k$ to converge its tail must converge to 0.

$$
lim_n to infty sum_k=n^inftyc_k = 0
$$

i.e. for every $epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$
left| sum_k=n^inftyc_k right| < epsilon
$$

We can prove that there exists a $epsilon$ such that for all $n$ the sum is bigger than $epsilon$ and thus the sum diverges.

$$
sum_k=n^inftyfraca_kb_k geq frac1b_nsum_k=n^inftya_k = 1 = epsilon
$$

First answer was wrong. Here is my new try :

If $undersetnto+inftylim fraca_nb_n$ is not $ 0 $ or does not exist, $sum fraca_nb_n$ is directly divergent.

Otherwise we have $undersetnto+inftylim fraca_nb_n = 0 $ so $b_n+1 sim b_n $ . It implies $sum fraca_nb_n = sum fracb_n-b_n+1b_n$ and $sum fracb_n-b_n+1b_n+1$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $nto+infty$, we can use an integral comparison :

$$ sum_n=1^N fracb_n-b_n+1b_n+1 ge sum_n=1^N int_b_n+1^b_n fracdxx = int_b_N+1^b_1 fracdxx = ln left(fracb_1b_N+1right) undersetNto+inftylongrightarrow+infty $$