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In a real analysis exam last year, I had to show if the following sequence $(u_n)_{n \in \mathbb{N}}$ defined by $u_n = \frac{2^n}{n + 2^n}$ is summable using property of sequences. However, I failed to justify correctly that it was not summable.

I tried to squeeze the sequence between $0$ and $1$ :

$$ 0 < \frac{2^n}{n + 2^n} < 1 $$

And because $(0)_{n \in \mathbb{N}}$ is summable but $(1)_{n \in \mathbb{N}}$ is not summable, I concluded that $\left(\frac{2^n}{n + 2^n}\right)_{n \in \mathbb{N}}$ is not summable.

I guess my squeezing is wrong. I was lately thinking about :

$$ 0 < \frac{2^n}{n + 2^n} < \frac{2^n}{n} $$

instead and I would be tempted to conclude that $\left( \frac{2^n}{n} \right)_{n \in \mathbb{N}}$ is not summable so $\left(\frac{2^n}{n + 2^n}\right)_{n \in \mathbb{N}}$ is not summable by comparison. Is it right ?

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  • $\begingroup$ you probably rememberred it wrong. $\endgroup$
    – faceclean
    May 16, 2020 at 20:03
  • $\begingroup$ $u_n=\frac{1}{\frac{n}{2^n}+1}>\frac{1}{1+n}$ $\endgroup$ May 16, 2020 at 20:06

4 Answers 4

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To show that $u_n$ is not summable you can use that $u_n = \frac{2^n}{n+2^n} = \frac{1}{n/2^n+1} \geq \frac{1}{n+1}$. $\frac{1}{n+1}$ is not summable (harmonic series).

You are only showing that $u_n < \frac{2^n}{n}$ and that the latter is not summable. This doesn't tell us anything about $u_n$ though.

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It is actually much simpler: as $n=o\bigl(2^n\bigr)$, we have that $n+2^n\sim_\infty 2^n$, hence the general term tends to $1$, and it should tend to $0$ if the series were convergent.

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You seem to be assuming that if a series is bounded above by one that isn't summable, then it isn't summable. But that obviously isn't true. For example, we have $0 < \frac{1}{n^2} < \frac{1}{n}$ and $\sum_{n=1}^{\infty}\frac{1}{n}$ isn't summable, but $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is. If you think about it, this should be obvious: if a series were bounded below by one that isn't summable, then it wouldn't be summable. Informally, if a series is larger than something that already sums to $\infty$, then it must itself sum to $\infty$. In contrast, if what you know is that the series is smaller than something that sums to $\infty$, that doesn't tell you anything.

Therefore, I will present a correct proof that the series isn't summable. We have \begin{align*}\frac{2^n}{n+2^n} &= \frac{1}{\frac{n}{2^n}+1} \\ &\geq \frac{1}{n+1}\end{align*} which isn't summable, so by what I said above, your series is also not summable.

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  • $\begingroup$ Thanks, but I checked in my lecture notes, it is said that let $v_n \leq u_n \leq w_n$, $\forall n \in \mathbb{N}$. if $v_n$ or $w_n$ are not summable we cannot say anything about the summability of $u_n$ $\endgroup$
    – Mathieu
    May 28, 2020 at 8:58
  • $\begingroup$ Your lecture notes are incomplete. If $0 < v_n \leq u_n$ for all $n$ and $v_n$ is not summable, then we can deduce that $u_n$ is not summable, using the Comparison Test. $\endgroup$
    – Prasiortle
    May 28, 2020 at 14:29
  • $\begingroup$ Allright, In fact you were right in a sense you were speaking about the convergence of the serie. I asked my teacher assistant and he pointed toward the fact that if $v_n \leq u_n$ and the sum of $v_n$ is divergent then the sum of $u_n$ is also divergent. Indeed, the sequence $\frac{1}{2^n}$ is summable but $-n < \frac{1}{2^n}$ and the sequence $-n$ is not summable. $\endgroup$
    – Mathieu
    May 30, 2020 at 17:53
  • $\begingroup$ That doesn't contradict what I said above, as $-n$ is not $>0$. $\endgroup$
    – Prasiortle
    May 31, 2020 at 4:46
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$u_n:=\dfrac{2^n}{n+2^n}\ge$

$ \dfrac{2^n}{2^n+2^n}= 1/2 >0;$

The series $\sum u_n$ diverges.

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