$a_n>0$, and the following attitude is true. Prove that the sequence is convergent.

$$a_n<a_{n-1}+\frac1{2^n},\quad n\ge2$$

So, what I tried here is to simply write down the elements of the sequence. If we have a "big enough" $n$, then $\frac12^n$ will converge to $0$, so what we have left is $a(n) < a(n-1)$ (the $n$-th element is smaller than the $n-1$-th element), which means that this sequence monotonically decreasing, after a big enough $n$.

Since this is a monotonically decreasing sequence, if we can find a number as a limit, then it must be convergent, but I can't find any.

Am I thinking wrong or what should be the correct way?

  • $\begingroup$ You also need the assumption that $a_n$ is increasing. Otherwise, $a_n = -n$ satisfies your assumption. $\endgroup$ Mar 2, 2015 at 16:23

2 Answers 2


An idea: I think you can prove that $(a_n)$ is cauchy by using the telescoping trick and the triangle inequality: For instance, let $m > n > 0$ and write

$$ | a_n - a_m | = |(a_n - a_{n-1}) + a_{n-1} - .... + (a_{m+1} - a_m) |$$

and use triangle inequality and we know you can make $\frac{1}{2^n} $ small as you want.

  • $\begingroup$ You might have a jump down that is larger than $1/2^k$ $\endgroup$
    – Empy2
    Mar 2, 2015 at 15:42
  • $\begingroup$ What shall I write with the trinagle inequality, I'm not quite sure. :) $\endgroup$
    – Atvin
    Mar 2, 2015 at 18:43

No, because your "reason" is the same if the $\frac1{2^n}$ is replaced by $\frac1{n}$, but in this case the $a_n$ can diverge.

You have to look at the particular bound. In this case, $a_n-a_{n-1} < \frac1{2^n} $.

If we sum this for a series of $n$, the left side telescopes like this: $(a_1-a_0)+(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+ ...+(a_n-a_{n-1}) = a_n-a_1 $. The right side is $\frac12 + \frac14 + \frac1{8} + \frac1{16}+ ... + \frac1{2^n} =1-\frac1{2^n} $.

Therefore $a_n-a_1 < 1-\frac1{2^n} < 1 $, so $a_n < 1+a_1$.

  • 1
    $\begingroup$ But that last part just shows the sequence is bounded -- to conclude, you need to elaborate a bit more (e.g., by showing that it can only have one adherence value, which then gives the result as it must have one by boundedness) $\endgroup$
    – Clement C.
    Mar 2, 2015 at 15:43
  • $\begingroup$ A bounded increasing sequence is convergent. You do not say that $a_n$ is increasing. If $a_n = -n$, then $a_n-a_{n-1} = -1 < 1/2^n$. $\endgroup$ Mar 2, 2015 at 15:51
  • $\begingroup$ Well, $a_n$ is positive, though. $\endgroup$
    – Clement C.
    Mar 2, 2015 at 17:30
  • $\begingroup$ True. I missed that. $\endgroup$ Mar 2, 2015 at 19:32

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