4
$\begingroup$

Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :

Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l>0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.

Am I correct by the following (sketch of) proof? :

1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| < \epsilon_1$ for all $n \ge N_1$.

2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|< \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.

3- Define $N = \max {\{N_1,N_2}\}$.

4- From Step (1), $a_{n+k} < (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} < (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})< \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.

5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' >0$ and repeat the process this time for a and b exchanged.

6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.

Thanks.

$\endgroup$
  • $\begingroup$ In step one do you mean $a_n/b_n$ instead of $a_k/b_k$? $\endgroup$ – Vim Mar 31 '17 at 23:46
  • $\begingroup$ Also, I think the logic in step4 is vague. It seems that you have defined the $\epsilon_3$ a posteriori rather than given it a priori. A proper argument should be: you first give $\epsilon_3$ for the criterion of convergence of $\sum b_k$, then proceed to find $N$'s. $\endgroup$ – Vim Mar 31 '17 at 23:55
  • $\begingroup$ @Vim, yes. You're right about the step 4. Thanks. $\endgroup$ – user231343 Apr 1 '17 at 22:27
  • $\begingroup$ Sorry for commenting late on this, but how does the convergence of $\sum_n a_n$ imply $|b_{n+1}+⋯+b_{n+k}|<\epsilon_1$? You're saying the convergence of $\sum_n a_n$ implies the convergence of $\sum_n b_n$, but you have yet to prove it? $\endgroup$ – Spencer Kraisler May 31 at 18:28
2
+50
$\begingroup$

Overall looks ok. But you can simplify the proof a lot just from the fact that $$\left|\frac{a_k}{b_k}-l\right|<\varepsilon \Leftrightarrow (l-\varepsilon)b_k<a_k<(l+\varepsilon)b_k, \forall k>N(\varepsilon)$$ Which means:

  1. If $\sum_{k=1}^{\infty}b_k<\infty$ then $$\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{N(\varepsilon)}a_k+\sum_{k=N(\varepsilon)+1}^{\infty}a_k<\sum_{k=1}^{N(\varepsilon)}a_k + (l+\varepsilon)\sum_{k=N(\varepsilon)+1}^{\infty}b_k<\infty$$

  2. If $\sum_{k=1}^{\infty}a_k<\infty$ then $$\sum_{k=1}^{\infty}b_k=\sum_{k=1}^{N(\varepsilon)}b_k+\sum_{k=N(\varepsilon)+1}^{\infty}b_k<\sum_{k=1}^{N(\varepsilon)}b_k + \frac{1}{l-\varepsilon}\sum_{k=N(\varepsilon)+1}^{\infty}a_k<\infty$$ $\varepsilon>0$ can be small enough so that $l-\varepsilon>0$.

$\endgroup$
  • $\begingroup$ Thanks a lot for the simple proof. $\endgroup$ – user231343 Apr 1 '17 at 22:27
3
$\begingroup$

As you wrote, both sequences $\dfrac{a_n}{b_n}, \dfrac{b_n}{a_n} $ converge. And convergent sequences are bounded. Thus both sequences $\dfrac{a_n}{b_n}, \dfrac{b_n}{a_n} $ are bounded, say by $A$ and $B$ respectively. This allows us to prove the result quickly: Suppose $\sum a_n$ converges. Since

$$b_n = \dfrac{b_n}{a_n}a_n \le Ba_n\,\,\text { for all }n,$$

we have for any $N,$

$$\sum_{n=1}^{N} b_n \le \sum_{n=1}^{N} Ba_n = B\sum_{n=1}^{N} a_n <B\sum_{n=1}^{\infty} a_n < \infty.$$

Thus the sequence of partial sums of $\sum b_n,$ which are monotonically increasing, are bounded above. Hence this sequence converges. By definition, that means $\sum_{n=1}^{\infty} b_n$ converges. Of course the proof the other way around is the same, and we're done.

$\endgroup$
0
$\begingroup$

Details, details.

In step 2- you should say " Suppose $\sum b_k$ converges ." You wrote $\sum a_k$ which I think is a typo. But do not say "Since $\sum b_k$ converges" as we don't know that it does. We are only trying, at this stage, to prove that convergence of $\sum b_k$ implies convergence of $\sum a_k.$..... And you have mentioned both $\epsilon_1$ and $\epsilon_2 .$ Possibly a typo . I suggest you eliminate $\epsilon_2$. Just use the same $\epsilon_1$ as in step 1-.

In step 4, the term $\epsilon_3$ appears for the first time, undefined. There should be a step 0- where we take an arbitrary $\epsilon_3>0.$ Then in step 1-, take $\epsilon_1>0$ small enough that $\epsilon_1(l+\epsilon_1)<\epsilon_3.$

Otherwise it's fine, clear and concise

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy