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Prove that if $a_n\gt 0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ diverges.

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This is the solution to this problem, but I'm having a hard time understanding it. Why does $a_k/(1+a_k)$ not converge to $0$ if $a_k$ doesn't converge to $0$?

I'd appreciate it if anyone could answer this question for me.

nomadicmathematician
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  • Write $\lim_{k->\infty}\frac{a_k}{1+a_k}=\lim_{k->\infty}\frac{1}{1+1/a_k}$

    what happens now if $a_k$ doesn't go to zero as $k -> \infty$?

     – tired Oct 06 '15 at 19:16
  • As @tired suggests, use the algebraic properties of the limit. If $\lim \limits_{k \to \infty} a_{k}$ exists, then $\lim \limits_{k \to \infty} 1 + a_{k}$ exists and is nonzero if $\lim \limits_{k \to \infty} a_{k} > 0$; but that implies $\lim \limits_{k \to \infty} \dfrac{a_k}{1 + a_k}$ exists, and the limit is $\dfrac{\lim \limits_{k \to \infty} a_{k}}{1 + \lim \limits_{k \to \infty} a_{k} }$, which is nonzero if $\lim \limits_{k \to \infty} a_{k} > 0$, since for any positive number $c$, $\dfrac{c}{1 + c}$ is not zero. – layman Oct 06 '15 at 19:19
  • But who says $\lim a_k$ has to exist? – Ted Shifrin Oct 06 '15 at 19:26
  • The proof has a TYPO in the last inequality. It should be $>$ and not $<$. So the last inequality should be $$ \sum_{k=1}^n\frac{a_k}{1+a_k} > \frac{ \sum_{k=1}^{n} a_k }{2} $$ Thus, the partial sums on the left are unbounded. – Ramiro Oct 06 '15 at 19:50
  • @TedShifrin If $\lim a_k$ does not exist or if $\lim a_k\neq 0$ then $\frac{a_k}{1+a_k}$ does not tend $0$ and the divergence of $\sum \frac{a_k}{1+a_k}$ is trivial. – Ramiro Oct 06 '15 at 20:14
  • Assuming the $\lim a_k=0$ (it is the only case less trivial), since we know that $a_k>0$ for all $k$, we may also correctly asssume WLOG that $0< a_k<1$ for all $k$. At tis point, as @TedShifrin has remarked, we can easily and very directly conclude that $\frac{a_k}{1+a_k}>\frac{a_k}{2}$. – Ramiro Oct 06 '15 at 20:28
  • https://math.stackexchange.com/q/131678/321264 – StubbornAtom May 03 '22 at 15:02

3 Answers3

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The solution is flawed. The conclusion in the final sentence does not follow from the preceding line (an upper bound won't do; we need a lower bound). But if we're assuming $0<a_k<1$, then $\dfrac{a_k}{1+a_k} > \dfrac{a_k}2$, and so we have bounded our series below by a divergent series.

The solution presented heads off in a far more complicated direction than is needed and then doesn't finish it off correctly.

By the way, if $a_k\ge 1$, then $\dfrac{a_k}{1+a_k}\ge \dfrac12$, and so infinitely many such terms will clearly give a divergent subseries.

Ted Shifrin
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  • The moral of the story is: Do not trust "solutions." If you must read them, read them critically and try to fix things that are wrong. – Ted Shifrin Oct 06 '15 at 19:39
  • If $a_k$ tends to $0$ only a finite number of $a_k$ will be greater or equal to $1$. – Ramiro Oct 06 '15 at 19:42
  • @Ramiro: Sure, but who on earth says that $a_k\to 0$? We are told that $\sum a_n$ diverges. – Ted Shifrin Oct 06 '15 at 19:43
  • Of course you have tacitly assumed that there exists a number, say $K$, such that either $a_k>1$ for all $k>K$ or $a_k\le 1$ for all $k>K$. It is possible that such a $K$ does not exist (e.g., $a_k=1+(-1)^n/2n$). This will not impact the result, but it is a case that seems to have been omitted. – Mark Viola Oct 06 '15 at 19:56
  • The first line of the solution presented in the question explains that if $a_k$ does not tend to $0$ then $\frac{a_k}{1+a_k}$ does not tend $0$ and the divergence will be "automatic".

    The case where the result is less trivial is exactly when $a_k \to 0$.

     – Ramiro Oct 06 '15 at 20:01
  • @Ramiro: Ah, my apologies. Although that first sentence does require some justification, so I was ignoring it. – Ted Shifrin Oct 06 '15 at 20:17
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I think there's just a typo in the solution; the final inequality should be $>$, not $<$. From the solution you gave, we have:

$$a_k - \frac{a_k}{1+a_k} < \frac{a_k}2$$

Now move the $\frac{a_k}{1+a_k}$ term to the right-hand side and the $a_k$ to the left to get:

$$\frac{a_k}2 < \frac{a_k}{1+a_k}$$

Taking partial sums, the left side is unbounded, therefore so is the right.

Théophile
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    Agreed. But this is approach unnecessarily overly complicated in the first place. – Ted Shifrin Oct 06 '15 at 19:37
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    @TedShifrin Indeed! Your solution is the natural way and is much clearer. My point is just that the book's solution works as written by flipping the last inequality, so it's not a fundamental flaw so much as a typo. – Théophile Oct 06 '15 at 19:38
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If $ v_k=\dfrac{a_k}{1 +a_k} $ then $a_k=\dfrac {v_k }{1-v_k } $

Then if $v_k $ converge to zero , $ a_k $ converge also to zero.

Hamou
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  • But $a_k\to 0$ does not mean that $\sum{a_k}$ converges. – Svetoslav Oct 06 '15 at 20:20
  • $a_k\to 0\iff v_k\to 0$ in this cas $a_k\sim v_k$, and $\sum a_k$ diverge $\iff $ $\sum v_k$ diverge. If $a_k$ does not converge to zero then $v_k$ does not converge to zero, and in this case $\sum v_k$ diverge. – Hamou Oct 06 '15 at 22:14