I am manipulating a series and have gotten this far:

$$ \lim_{n\to\infty} \sum_{m=1}^n \frac {(\frac{m}{n})^{p-1}}{1+ (\frac{m}{n})^p} \frac{1}{n}$$

I want to now say that this is a Riemann sum, which corresponds to the Riemann integral

$$ \int_0^1 \frac {x^{p-1}}{1+x^p}dx$$

which evaluates to ln(2)/p -- the answer I am looking for.

Have I converted to the integral correctly?

My concerns are: I just relabeled the discrete variable ($m/n$) as a continuous variable $x$. Also I converted the $1/n$ interval length to $dx$ -- this part I am ok with, just from checking the definition of Riemann integral and knowing that I have to let mesh($p$) = supremum of sub-interval lengths go to zero.

So, I sort of did it blindly but I would like to know what are the rules that I may have missed? E.g., does the variable ($m/n$) also have some size limits to adhere to? Does ($m/n$) have to be small?


EDIT: The main concern really is: in past problems, I converted to Riemann integrals that did not correspond to the Riemann sums that I was working with. An MSE contributor gave me the heads up -- and showed that my corresponding "Riemann integrals" actually had values different from the sums. But I used the same "rules" as I did with this current problem statement. Apparently, it was incorrect. So, I just want to know whether I have converted this sum to an integral correctly.

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    $\begingroup$ Note that $\displaystyle lim_{n->\infty}$ is not a conventionally correct way of typesetting this notation, but $\displaystyle \lim_{n\to\infty}$ is. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 6 '15 at 1:13
  • $\begingroup$ Ok, got it - I will keep that in mind. Thanks @MichaelHardy. $\endgroup$ – User001 Oct 6 '15 at 1:15
  • $\begingroup$ looks fine to me $\endgroup$ – steven gregory Oct 6 '15 at 1:20
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    $\begingroup$ Looks fine. Was the other integral improper? That could make thing complicate. $\endgroup$ – user251257 Oct 6 '15 at 1:25
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    $\begingroup$ If you integrate over $[a, b]$, you just need a sequence of partitions of $[a,b]$ that get arbitrarily fine (like in the definition of Riemann integral). $\endgroup$ – user251257 Oct 6 '15 at 1:40

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