Here's Prob. 11 (d), Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $a_n > 0$ and that the series $\sum a_n$ is divergent. Then what can be said about the convergence of $$\sum \frac{a_n}{1+na_n}?$$

I know that if $\left\{ n a_n \right\}$ is bounded above or has a positive lower bound, then this series diverges. Does the converse hold as well?

Suppose that $\left\{ n a_n \right\}$ is neither bounded above nor has a positive lower bound. Then there is a subsequence $\left\{ n_k a_{n_k} \right\}$ such that $$n_k a_{n_k} \geq k$$ for all $k$.

And, there is a subsequence $\left\{ m_r a_{m_r} \right\}$ such that $$m_r a_{m_r} < \frac{1}{r}$$ for all $r$.

What next?

  • $\begingroup$ for $a_n = 1/n^2$, $n a_n$ is bounded from above but the series converges. Perhaps you misphrased (or I misunderstood)? $\endgroup$ – spaceisdarkgreen Dec 31 '16 at 8:53
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    $\begingroup$ For $\sum a_n$ divergent: math.stackexchange.com/questions/1990396/…. $\endgroup$ – Fernando Revilla Dec 31 '16 at 8:59
  • $\begingroup$ If you take $a_n = 1/n$ then you have $\sum \frac{a_n}{1+a_n} = \sum \frac{1}{2n}$ which diverges. $\endgroup$ – na1201 Dec 31 '16 at 9:01
  • $\begingroup$ @manhattan but you see, if $a_n = \frac{1}{n}$, then $\left\{ n a_n \right\}$ is still bounded above and has a positive lower bound. $\endgroup$ – Saaqib Mahmood Dec 31 '16 at 9:18
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    $\begingroup$ If $a_n = n^{2\cdot(-1)^n-1}$, then $\{na_n\}$ is neither bounded from above nor bounded from below by a positive number, but $\sum{\frac{a_n}{1+na_n}}$ still diverges since it is at least the sum of $\frac{n}{1+n^2}$ over all even integers $n$. $\endgroup$ – Joey Zou Dec 31 '16 at 9:52
  • Answering the Baby Rudin question:

You cannot say anything.

It may be divergent: for instance, take $(a_n)_{n\geq 1}$ to be identically $1$, or even $(a_n)_{n\geq 1}$ to be defined by $a_n = \frac{1}{n}$ (two natural examples).

It may be convergent: for instance, $a_n = \begin{cases} 1 &\text{ if } n=2^k \text{ for some }k\geq 0\\ 0 &\text{ otherwise}\end{cases}$ (you can replace $0$ by $2^{-n}$ if you want to enforce that the sequence be positive). Since $(a_n)_n$ does not converge to $0$, clearly the series $\sum_n a_n$ diverges. Yet, $\sum_{n=1}^\infty \frac{a_n}{1+n a_n} = \sum_{k=0}^\infty \frac{1}{1+2^k} < \infty$.

  • Answering the OP's followup question:

Take $(a_n)_{n\geq 1}$ defined by $$ a_n = \begin{cases} 2^n & \text{ for even } n\\ \frac{1}{2^n} & \text{ otherwise.} \end{cases} $$ Then $\sum_{n=1}^\infty a_n$ clearly diverges, and so does $\sum_{n=1}^\infty \frac{a_n}{1+n a_n} \geq \sum_{n=1}^\infty \frac{a_{2n}}{1+2n a_{2n}} = \sum_{n=1}^\infty \frac{1}{\frac{1}{2^{2n}}+2n}$. But $(a_n)_{n\geq 1}$ has neither a finite upper bound nor a positive lower bound.

  • $\begingroup$ (I may have misunderstood the question. This is answering the baby Rudin problem. Let me know if this is off topic) $\endgroup$ – Clement C. Dec 31 '16 at 10:40
  • $\begingroup$ thank you for your answer, but my question was, if (i) $a_n > 0$, (ii) if $\sum a_n $ is divergent, and (iii) if $\sum \frac{a_n}{1+na_n}$ is divergent, then can we prove that the sequence $\left\{ na_n \right\}$ is either bounded from above or has a positive lower bound? $\endgroup$ – Saaqib Mahmood Dec 31 '16 at 10:43
  • $\begingroup$ In your first examples, $na_n$ has a positive lower bound. The OP asked what happens when $ na_n$ is unbounded above and has no positive lower bound..... If $a_n=2^{-n}$ when $n$ is odd, and $a_n=1/\sqrt n$ when $n$ is even then $\sum_na_n/(1+na_n)$ diverges. .... I like your convergent example. $\endgroup$ – DanielWainfleet Dec 31 '16 at 10:47
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    $\begingroup$ Yes, I realized after posting (slow connection) that I addressed the Baby Rudin question, not the OP's. But I still must be missing something: what about $(a_n)$ defined by $a_n = 2^n$ for $n$ even and $2^{-n}$ for $n$ odd, then? Both $\sum_n a_n$ and $\sum_n \frac{a_n}{1+ n a_n}$ diverge, and there is no positive lower bound nor upper bound to $(a_n)$. @SaaqibMahmuud $\endgroup$ – Clement C. Dec 31 '16 at 10:57
  • $\begingroup$ (after reading the second comment, this is basically the same counterexample as @user254665 's ) $\endgroup$ – Clement C. Dec 31 '16 at 10:59

This one is super hard but this is what I have; the claims can be proven but I didn't do that here. Consider this to be messy work instead:

$\textbf{Claim 1}$: If inf $\left\{a_{n}\right\}=0$, then $\left\{a_{n}\right\}$ has a subsequence that converges to $0$.

$\textbf{Claim 2}$: If $a_{n}\longrightarrow 0$, then $\left\{a_{n}\right\}$ has a decreasing subsequence.

$\textbf{Claim 3}$: $s_{n}=a_{1}+\cdots+a_{n}$ forms a divergent sequence of partial sums, where $a_{n}>0$, then all of it's subsequences are unbounded.

Now, if $\sum a_{n}$ is divergent with inf $\left\{a_{n}\right\}=0, a_{n}>0$, then it has a subsequence that converges to $0$ which in turn has a decreasing subsequence. There is $N$ such that if $n\geq N, s_{n}>1$. If $\left\{s_{n_{k}}\right\}$ is the sequence of partial sums whose last term is the smallest (there is such a sequence because we have a decreasing sequence of terms), then for $n_{k}\geq N, \displaystyle \frac{a_{n_{k}}}{1+n_{k}a_{n_{k}}}\geq \frac{a_{n_{k}}}{1+s_{n_{k}}}>\frac{a_{n_{k}}}{2s_{n_{k}}}$. By part(b) and claim 3, the right-most expression has a divergent series and therefore

$\displaystyle \left\{\frac{a_{n_{k}}}{1+n_{k}a_{n_{k}}}\right\}$ has a divergent series. But this series is a subsequence of the partial sums

of $\displaystyle \sum \frac{a_{n}}{1+na_{n}}$, and thus, the original series diverges.

If $inf \left\{a_{n}\right\}=c>0$ and $a_{n}<1\forall n\Rightarrow \displaystyle \frac{a_{n}}{1+na_{n}}\geq \frac{c}{1+n}$ which forms a divergent series. Finally, if there is a subsequence with $a_{n_{k}}>1\Rightarrow \displaystyle \frac{a_{n_{k}}}{1+na_{n_{k}}}>\frac{a_{n_{k}}}{a_{n_{k}}+na_{n_{k}}}=\frac{1}{1+n}$.


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