# Is the sequence $u_n = \frac{2^n}{n + 2^n}$ summable?

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 ?

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

## 4 Answers

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.

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.

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.

• 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 $u_n$ May 28, 2020 at 8:58
• 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. May 28, 2020 at 14:29
• 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. May 30, 2020 at 17:53
• That doesn't contradict what I said above, as $-n$ is not $>0$. May 31, 2020 at 4:46

$$u_n:=\dfrac{2^n}{n+2^n}\ge$$

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

The series $$\sum u_n$$ diverges.