1
$\begingroup$

Let $(a_k)_{k \geq 1}$ and $(b_k)_{k \geq 1}$ be two strictly positive sequences such that $$\frac{a_{k+1}}{a_k} \leq \frac{b_{k+1}}{b_k}$$ I have to prove that if $\sum_{k=1}^\infty b_k$ converges, then $\sum_{k=1}^\infty a_k$ must converges as well.

I found that $a_k \leq C b_k$ for $C > 0$. So if $\sum_{k=1}^\infty b_k$ converge, then $\sum_{k=1}^\infty C b_k$ converges as well. It is not clear for me that $\sum_{k=1}^\infty a_k$ must converges using the squeeze theorem. The best I have so far is $$0 \leq \sum_{k=1}^n a_k \leq \sum_{k=1}^n C b_k$$ So $$0 \leq \sum_{k=1}^\infty a_k \leq \sum_{k=1}^\infty C b_k = L \in \mathbb{R}$$

But nothing specify $L=0$. Am I missing something?

$\endgroup$
2
  • 2
    $\begingroup$ Why would you expect $L$ to be $0$? It can't be $0$, because then the series would converge to $0$, which is impossible if all terms are positive. What you want to use here is that the partial sums form a strictly increasing sequence, so the series converges if and only if the partial sums are bounded above. Not sure what "Squeeze Theorem" you are trying to use here, given that the information you have does not let you determine the actual value of $\sum a_n$. So what would you squeeze it with? $\endgroup$ Mar 1 '21 at 20:44
1
$\begingroup$

Let $A_n := \sum_{k=1}^na_k$ and $B_n := \sum_{k=1}^nb_k$. Since $(B_n)$ is Cauchy, for $\varepsilon>0$ there exists $N$ such that $|B_n - B_m|<\frac{\varepsilon}{C}$ for $n\ge m\ge N$. Hence, for $n\ge m$ we obtain $$ |A_n-A_m| = \left|\sum_{k=1}^na_k - \sum_{k=1}^ma_k\right| = \sum_{k=m+1}^na_k\le C\sum_{k=m+1}^nb_k = C|B_n-B_m| < \varepsilon. $$ Therefore, $(A_n)$ is a Cauchy sequence and thus converges.

$\endgroup$
0
1
$\begingroup$

Well, if we have $$\frac{a_{k+1}}{a_k} \leq \frac{b_{k+1}}{b_k},$$ we can prove by induction $$a_k\le\frac{a_1}{b_1}\,b_k$$ for all $k\ge1$, and the rest is an easy consequence.

$\endgroup$
1
  • 2
    $\begingroup$ They already wrote they have such an inequality. $\endgroup$
    – amsmath
    Mar 1 '21 at 20:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.