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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?

EDIT :

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!

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Are you sure it says 'sequences' and not 'series'? $\endgroup$ – filterjuice Sep 14 '15 at 3:16
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $$a_n=\sum_{k=1}^n\frac1{2^k+k} =\frac1{2^1+1}+\frac1{2^2+2}+\frac1{2^3+3}+\cdots+\frac1{2^n+n}\ .$$

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