3

This is from an MCQ contest.

Let $\sum\limits_{n\geq 1}a_n$ be a convergent series of positive terms. Which of the following hold?

  • $1]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ and $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ are convergent.
  • $2]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ and $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ are divergent.
  • $3]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ is Divergent and $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ is convergent
  • $4]$ none of the previous statements is correct

i come up with this conter example :

  • $1]$ $\sum_{n\geq 1} \dfrac{1}{n^2}$ convergent so let 's verify: $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}=\sum\limits_{n\geq 1}\dfrac{n^2}{1+n^{2}}$

or $\lim_{n\to +\infty }\dfrac{n^2}{1+n^{2}}=1 \neq 0$ then $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ divergent thus $1]$ False i don't need to verify $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ because there is 'and' in the statement $1]$

  • $2]$ in this case i need to check the nature of $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$

$\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}=\sum\limits_{n\geq 1}\dfrac{1}{1+n^2}$ i don't know to calculate i use just Wolframe and it's convergent

  • $3]$ i can't say that statement is true just becuase one example work i need to prove it but i don't now how

  • Is my proof correct

  • Is there any kind of reasoning one may use during a contest, when a quick answer is needed
Did
  • 279,727
Educ
  • 4,780

2 Answers2

2

From your solution of (1), I'll assume that $\sum a_n$ is convergent. If so, your argument for (1) is correct. For (2) and (3) (and also (1)) note that

  1. $\displaystyle\frac{a_n}{a_n+1}<a_n$ so $\sum\dfrac{a_n}{a_n+1}$ is convergent.
  2. $\dfrac{1}{a_n+1}\to 1$ as $n\to \infty$ so $\sum\frac{1}{a_n+1}$ is divergent.

I hope that also answers your last question. Though honestly I'd say there's no magic recipe in handling convergence of a series.

Quang Hoang
  • 15,854
  • how can i prove that $\sum \dfrac{1}{a_n+1}$ convergent if $\sum a_n$ i can't prove it based in exampl – Educ Sep 21 '15 at 22:27
  • That's the Comparison Test, one of the basic, and probably the most commonly used, tests for convergence: A positive series bounded above by a convergent series is convergent. – Quang Hoang Sep 21 '15 at 22:38
  • you use that for $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ not for $\sum \dfrac{1}{a_n+1}$ i'm talking about the latter – Educ Sep 21 '15 at 22:41
  • 1
    Ah, I saw convergent. The series $\sum\dfrac{1}{a_n+1}$ is divergent. Exactly as you argued in your question ($\sum a_n$ is convergent, so $a_n\to 0$). – Quang Hoang Sep 21 '15 at 22:48
2

Much of the analysis in the OP was fine. To add some thoughts, recall that if a series converges, then it's terms must approach zero as $n\to 0$.

Therefore, if $\sum_{n=1}^{\infty}a_n$ converges, then $\lim_{n\to \infty}a_n=0$. This implies that $\lim_{n\to \infty}\frac{1}{1+a_n}=1$, which in turn implies that $\sum_{n=1}^{\infty}\frac{1}{1+a_n}$ diverges.

Now, since $a_n\to 0$, then there exists a number $N$ so that $a_n<1$ whenever $n>N$. Then, for such an $N$, $\frac{a_n}{1+a_n}<\frac12 a_n$. Inasmuch as the series $\sum_{n=1}^{\infty}\frac12 a_n$ converges, then so does the series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$.

The answer is, therefore, number $3$.

Mark Viola
  • 179,405