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Let $$f(x)=\frac{e^{2x-1}}{1+e^{2x-1}}$$ Then find $$\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)$$

How I proceed: $$\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=\int_{1}^{1233}\frac{e^{\frac{2x}{1234}-1}}{1+e^{\frac{2x}{1234}-1}}~dx$$ then how I solve this integral. please help.

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Why should the sum be equal to the integral –  Amr Dec 22 '12 at 11:43
    
Amr is spot on. You are looking at this in a wrong way. –  Nameless Dec 22 '12 at 11:45
    
Then how I proceed? –  Argha Dec 22 '12 at 11:46

3 Answers 3

up vote 4 down vote accepted

Note that:

\begin{align} f(k/1234)+f((1234-k)/1234)=& \frac{e^{\frac{k}{617}-1}}{e^{\frac{k}{617}-1}+1}+\frac{e^{\frac{1234-k}{617}-1}}{e^{\frac{1234-k}{617}-1}+1}&=&\\ =&\frac{e^{k/617}}{e^{k/617}+e}+\frac{e}{e^{k/617}+e}&=&1 \end{align}

So your sum is equal to $616+f(1/2)$.

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This is nice +1 –  Amr Dec 22 '12 at 12:06
    
Really it is good.Thank you. –  Argha Dec 22 '12 at 12:10
    
Thanks to Mathematica for simplifying the terms, I wouldn't have voluntarily done this. –  akkkk Dec 22 '12 at 12:11

Observe that $f(x)+f(1-x)=1$

Put $x=\frac k{1234}\implies f\left(\frac k{1234}\right)+ f\left(\frac {1234-k}{1234}\right)=1$

Then put $k=1,2,,\cdots ,\frac{1234}2=617$

so $ f\left(\frac 1{1234}\right)+ f\left(\frac {1234-1}{1234}\right)=1,$

$ f\left(\frac 2{1234}\right)+ f\left(\frac {1234-2}{1234}\right)=1,$

**

$ f\left(\frac {617}{1234}\right)+ f\left(\frac {1234-617}{1234}\right)=1\implies f\left(\frac 12\right)=\frac12$

Adding we get, $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=617(1)-f\left(\frac {617}{1234}\right)=617-\frac12$

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Thank you for giving me complete answer. –  Argha Dec 22 '12 at 12:12
1  
@Argha, thanks for putting up a nice problem. –  lab bhattacharjee Dec 22 '12 at 12:14
    
I got this problem and stuck. So I put it. But thank you for the comment. –  Argha Dec 22 '12 at 12:19
    
@Argha, welcome. –  lab bhattacharjee Dec 22 '12 at 12:19
    
I like how you can solve for $f(1/2)$. –  akkkk Dec 22 '12 at 12:49

I don't know if a closed form for your sum exists. However, we note that:

$$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{k=1}^{N}f(\frac{k}{N})=\int_0^1f(x)dx$$

Thus, for large values of $N$ you can estimate the sum using the integral.

To evaluate the integral, we note that $\frac{d}{dx}[e^{2x-1}+1]=2e^{2x-1}$, thus: $$\int_0^1\frac{e^{2x-1}}{e^{2x-1}+1}dx=\frac{1}{2}\int_0^1\frac{2e^{2x-1}}{e^{2x-1}+1}dx=[\frac{1}{2}\log(1+e^{2x-1})]_{x=0}^{x=1}$$

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Yes I will add it –  Amr Dec 22 '12 at 11:57
    
The value of the integral ? –  Amr Dec 22 '12 at 12:00
    
@Argha See akkkk's answer,way better than mine –  Amr Dec 22 '12 at 12:07

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