6

Find the sum of $n$ terms the following series:

$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$

$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$

But after that I'm not sure as to how to proceed

Michael Hardy
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Rohan
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4 Answers4

12

Hint: For $x\ne 1$, add $\frac{-1}{x-1}$ in front, and observe the telescoping.

André Nicolas
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  • That's somewhat magical; Thank you. – Rohan Oct 28 '14 at 17:50
  • How did you understand that we have to add this? – Rohan Oct 28 '14 at 17:55
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    You are welcome. There are less magical ways of doing it, but they take longer, and I got lazy. – André Nicolas Oct 28 '14 at 17:56
  • Telescoping occurs surprisingly often, it can be considered more of a tool than a trick. As to how I got to it, it is largely a matter of having used analogous tricks before. – André Nicolas Oct 28 '14 at 17:59
  • Could you please suggest some good books or a good websites on maths :) , Would like to learn, practice and become a pro like you – Rohan Oct 28 '14 at 18:01
  • It depends on whether at this stage you are contest oriented or general mathematics oriented. For contest-oriented, the web site the Art of Problem solving is good. There are also many contest-oriented books, at various levels. At Olympiad level, Engel, Problem solving Strategies is good. There are others, one by Zeitz. For general mathematics, it is more profitable to study subjects, whatever appeals to you. – André Nicolas Oct 28 '14 at 18:07
5

Just another approach.

Let $x$ be a real number such that $|x|>1$. Observe that $$ 1+x^{2^{n}}=\frac{x^{2^{n+1}}-1}{x^{2^{n}}-1} \quad n=0,1,2,\ldots, $$ giving $$ \prod_{n=0}^{N} \left(1+x^{2^{n}}\right)=\frac{x^{2^{N+1}}-1}{x-1} $$ Applying the logarithmic function gives $$ \sum_{n=0}^{N}\log \left(1+x^{2^{n}}\right)=\log\left(x^{2^{N+1}}-1\right)-\log (x-1) $$ differentiating with respect to $x$ $$ \sum_{n=0}^{N}\frac{2^{n}x^{2^{n}-1}}{x^{2^{n}}+1}=\frac{2^{N+1}x^{2^{N+1}-1}}{x^{2^{N+1}}-1}-\frac{1}{x-1} $$ equivalently $$ \sum_{n=0}^{N}\frac{2^{n}}{x^{2^{n}}+1}=\frac{1}{x-1}-\frac{2^{N+1}}{x^{2^{N+1}}-1} $$ and then making $N$ tend to $+\infty$ leads to a closed form for your series.

Olivier Oloa
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3

Hint: Try some small cases ($n=1,2,3$). The pattern should become clear, and you can most likely prove it by induction on $n$.

rogerl
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  • I think that Mathematical Induction would be useful only in proving a result ie when we know the answer already, Please correct me if I am wrong. – Rohan Oct 28 '14 at 18:04
  • @Rohan conjecture an answer and apply induction on it. – UserX Oct 28 '14 at 18:42
1

note: having just seen Andre's lovely solution this is very much a hammer-and-chisel job! however there was some devil in the detail, and i did learn something from the exercise!

define: $$ D_m=\prod_{k=1}^m (1+x^{2^{m-1}}) = \sum_{k=0}^{2^m-1}x^k $$ and

$$ N_m = \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} \tag{1} $$ and now define the sum $$ S_m=\frac{N_m}{D_m} $$ then $$ S_m+\frac{2^m}{x^{2^m}+1} = \frac{(x^{2^m}+1)N_m + 2^mD_m}{D_{m+1}} $$

from (1) we have $$ N_{m+1} = \sum_{k=1}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\ = \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} +\sum_{k=1}^{2^m-1} 2^m x^{k-1} + \sum_{k=2^m}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\ = N_m + 2^m(D_m -x^{2^m-1}) + \sum_{k=0}^{2^m-1} (2^m-k)x^{2^m+k-1} \\ = (1+x^{2^m}) N_m + 2^m D_m $$ hence, with the obvious definition of $S_m$ we have the inductive step: $$ S_m = \frac{N_m}{D_m} \Rightarrow S_{m+1} = \frac{N_{m+1}}{D_{m+1}} $$

David Holden
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