$\sum a_{n}=\infty$. P T there is $b_{n}$ such that $\lim_{n \rightarrow \infty} b_{n}=0$ and $\sum a_{n} b_{n}= \infty$.

Question

Prove that there exists a sequence of positive real numbers $b_{n}$ such that $\lim_{n \rightarrow \infty}=0$ and $\sum a_{n} b_{n}= \infty$.

My attempt:

If possible suppose that there is a sequence $b_{n}$ convergese to 0 and $\sum a_{n} b_{n}$ is also convergent.

This implies, given $\epsilon>0$, there is a positive integer N such that $\sum_{n=p}^{n=q} a_{n} b_{n} <\epsilon$ for $q\geq p\geq N$.

Summation by parts $|\sum_{n=p}^{n=q} a_{n} b_{n}|$=$|\sum_{n=p}^{n=q-1} A_{n} (b_{n}-b_{n+1})+A_{q}b_{q}-A_{p-1}b_{p}|< \epsilon$. We know that $b_{n}$ is bounded, so it is bounded by K.

Now, $|\sum_{n=p}^{n=q} a_{n} b_{n}|=K|\sum_{n=p}^{n=q-1} A_{n} +A_{q}-A_{p-1}|< \epsilon$.

This implies $|\sum_{n=p}^{n=q} A_{n} +A_{q}-A_{p-1}|< \frac{\epsilon}{K}$. This is a contradiction because of $\sum a_{n}=\infty$.

Is there is any mistake in this proof?

Answer 1

Hint: We can construct a suitable sequence $b_n$ as follows: first, define a sequence of $k_m$ for $m = 0,1,2,\dots$ that satisfies

• $k_0 = 0$
• $\sum_{j={k_{n-1}}}^{k_{n}} a_n \geq m$ for $m = 1,2,\dots$.

Then, define the sequence $b_n$ by $b_n = \frac 1{m(n)}$, where $$ m(n) = \max\{m \in \Bbb N : n \geq k_m\}. $$