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Q. Find the value of - $$ \lim_{n\to\infty} \sum_{r=0}^{n} \frac{2^r}{5^{2^r} +1} $$

My attempt - I seem to be clueless to this problem. Though I think that later terms would be much small and negligible( Imagine how large would be $ 5^{2^r} $ after 3-4 terms), so I calculated the sum of first 3-4 terms and my answer was just around the actual answer ( just a difference of $0.002$ ). But I wonder if there is an actual method to solve this problem? If it is, would you please share it to me?

Any help would be appreciated

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  • $\begingroup$ Hint: Comparison test. Further hint: $0.25$ $\endgroup$ – Joseph Eck Jun 8 '18 at 7:35
  • $\begingroup$ @Joseph Eck Sorry, I don't think I got you! $\endgroup$ – Creep Anonymous Jun 8 '18 at 7:37
  • $\begingroup$ I’d suggest checking out material like this $\endgroup$ – Joseph Eck Jun 8 '18 at 7:39
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    $\begingroup$ @JosephEck - the comparison test is useful to show whether or not a series converges. It does not help in finding the value of the series, which is the problem here. $\endgroup$ – Paul Sinclair Jun 8 '18 at 16:56
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    $\begingroup$ Related, but more general: math.stackexchange.com/questions/2731886/… $\endgroup$ – Misha Lavrov Jun 8 '18 at 20:11
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Recall every natural number $m$ can be uniquely decomposed into a sum of powers of $2$, $$m = \sum_{r=0}^\infty b_r 2^r\quad\text{ where }\quad b_r \in \{ 0, 1 \}$$

For any $x \in (0,1)$, this leads to

$$\prod_{r=0}^\infty \left(1 + x^{2^r}\right) = \sum_{m=0}^\infty x^m = \frac{1}{1-x}$$ Taking logarithm on both sides and apply $x\frac{d}{dx}$ to them, we obtain

$$\sum_{r=0}^\infty \frac{2^r x^{2^r}}{1 + x^{2^r}} = \frac{x}{1-x}$$

Substitute $x$ by $\frac15$, we find

$$\sum_{r=0}^\infty \frac{2^r}{5^{2^r} + 1} = \sum_{r=0}^\infty \frac{2^r\left(\frac15\right)^{2^r}}{1 + \left(\frac15\right)^{2^r}} = \frac{\frac15}{1-\frac15} = \frac14$$

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    $\begingroup$ Could you elaborate on how you get the equality between the product and the sum? $\endgroup$ – Ant Jun 8 '18 at 16:20
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    $\begingroup$ @Ant, Notice $$\prod_{r=0}^{N-1} (1 + x^{2^r}) = \prod_{r=0}^{N-1} \sum_{b_r=0}^1 x^{b_r2^r} = \sum_{(b_0,b_1,\ldots,b_{N-1}) \in \{0,1\}^N} \prod_{r=0}^N x^{b_r 2^r} = \sum_{(b_0,b_1,\ldots,b_{N-1}) \in \{0,1\}^N} x^{\sum_{r=0}^N b_r 2^r} $$ Treat $(b_0,b_1,\ldots,b_{N-1})$ as the binary representation of a number with at most $N$ binary digits. When $(b_0,\ldots,b_{N-1})$ runs over all possible combinations in $\{0,1\}^N$, $\sum_{r=0}^N b_r 2^r$ will run over every natural number $m \le 2^N - 1$ once and only once. So the last sum equals to $\sum_{m=0}^{2^N - 1} x^m$ $\endgroup$ – achille hui Jun 8 '18 at 16:41
  • $\begingroup$ Wow! That's a very nice trick!! :D Last thing I am confused is how you exchange the sum and the product - the rest is clear. Thank you, and very nice method! $\endgroup$ – Ant Jun 8 '18 at 16:53
  • $\begingroup$ @Ant which one? do you mean the second '=' in comment. you just need to expand a few by yourselves (say for N = 2 and 3) and you will see the pattern. $\endgroup$ – achille hui Jun 8 '18 at 17:01
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Another user pisco originally posted this, but kept deleting his/her answer. It however is completely elementary and hence in my opinion superior to using power series and termwise differentiation (which require non-trivial theorems):

$$\frac{2^r}{5^{2^r}+1} = \frac{2^r}{5^{2^r}-1} - \frac{2^{r+1}}{5^{2^{r+1}}-1}.$$

Then just take $\displaystyle\sum_{r=0}^n$ of both sides and observe that the 'remainder' term vanishes as $n \to \infty$.

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  • $\begingroup$ Of course! It's an official approach and easy to understand! Where terms get cancelled $\endgroup$ – Creep Anonymous Jun 8 '18 at 15:23
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    $\begingroup$ Hello user21820! How did you know to decompose it in such a way? Were you trying to solve it by telescoping and actively looking for ways to decompose it? $\endgroup$ – Ovi Jun 8 '18 at 23:33
  • $\begingroup$ @Ovi You just have to multiply $5^{2^r}-1$ to numerator and denominator and then you will find it yourself! I don't think we have several ways to solve a problem like this, so only decomposing it remains the way or solve by higher tricks as achille hui did it! $\endgroup$ – Creep Anonymous Jun 9 '18 at 4:01
  • $\begingroup$ @CreepAnonymous Ah thank you very much! $\endgroup$ – Ovi Jun 9 '18 at 4:19
  • $\begingroup$ @Ovi: Yes indeed I was both looking for a telescoping (because of this) and also the identity "$(x+y)(x-y) = x^2-y^2$" and its relatives so frequently show up in algebraic identities that one 'must' try them. $\endgroup$ – user21820 Jun 9 '18 at 6:03

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