I have the following series: $$a_n=\frac{1}{2^n+n}$$

For some reason in the answer it says it increasing, I'm trying to prove it by induction but $$a_1=\frac{1}{3} \geq \frac{1}{6}=a_2$$

and for increasing we should get that $a_n \leq a_{n+1}$ so I think it should be decreasing instead of increasing.

Is there a problem with the question? Is the series actually increasing?


This is the book proof :

$$a_{n+1}-a_n=(\frac{1}{2+1}+\frac{1}{2^2+2}+...+\frac{1}{2^n+n}+\frac{1}{2^{n+1}+n+1})-(\frac{1}{2+1}+\frac{1}{2^2+2}+...+\frac{1}{2^n+n})=\frac{1}{2^{n+1}+n+1} \geq 0$$

I find it to be a mistake since $$a_{n+1}=\frac{1}{2^{n+1}+n+1}$$ Therefore $\frac{1}{2+1} \not\in a_{n+1}$

Thank you!

  • $\begingroup$ It is decreasing. Differentiate it. Maybe your text is saying that the sum is increasing I guess? $\endgroup$ – Rubertos Sep 14 '15 at 2:55
  • $\begingroup$ It’s clearly decreasing, since the numerator is constant and the denominator is increasing. $\endgroup$ – Brian M. Scott Sep 14 '15 at 2:55
  • $\begingroup$ @Rubertos I need to show that $\{a_n\}$ convergences, the question gives a hint to show its increasing and bounded. $\endgroup$ – JaVaPG Sep 14 '15 at 2:59
  • 2
    $\begingroup$ It looks like the book is talking about the partial sums $A_n=\sum_{k=1}^na_k$ rather than the $a_n$ themselves. $\endgroup$ – Milo Brandt Sep 14 '15 at 3:13
  • 1
    $\begingroup$ Are you sure it says 'sequences' and not 'series'? $\endgroup$ – filterjuice Sep 14 '15 at 3:16

I think your confusion lies in the difference between sequences and series. Before your edit from 'the following sequence' to 'the following series' you were correct. But a series is defined as the partial sum $$a_n = \sum_{i=1}^na_i$$ where $a_i$ is somewhat confusingly the $i^{th}$ component of the sequence $a_n$.

So in your case $$a_n = \sum_{i=1}^n\frac{1}{2^i+i}$$ which is clearly increasing as $n$ increases since the denominator is positive i.e. you're always adding on a positive number.

  • $\begingroup$ Thank you for your answer, So let me understand if I got it right, they defined $$a_n=\frac{1}{2+1}+\frac{1}{2^2+2}+...+\frac{1}{2^n+n}$$ which is $$a_n = \sum_{i=1}^n\frac{1}{2^i+i}$$, so If I show that the sequence is raising therefore, $\{a_n\}$ (series) is raising and is clear that it raising since we always add a positive number, also we know that $$$a_{n+1} = \sum_{i=1}^n\frac{1}{2^{i+1}+i+1}$$, but I still think that the proof in book is wrong since since $\frac{1}{2+1} \not\in a_{n+1}$ can you please explain the book's proof? $\endgroup$ – JaVaPG Sep 14 '15 at 14:13
  • 1
    $\begingroup$ Actually $\frac{1}{2+1} \in a_{n+1}$ because you have written the series $a_{n+1}$ incorrectly: the only thing that should change is the index you are summing to: $$\{a_{n+1}\}=\sum_{i=1}^{n+1} \frac{1}{2^i+i}$$ So in fact that first element is there. $\endgroup$ – filterjuice Sep 14 '15 at 15:04

It's clear from the working that the question has been stated incorrectly in the book (or that you have misunderstood it). It's not $$a_n=\frac1{2^n+n}$$ but $$a_n=\sum_{k=1}^n\frac1{2^k+k} =\frac1{2^1+1}+\frac1{2^2+2}+\frac1{2^3+3}+\cdots+\frac1{2^n+n}\ .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.