Hi I was wondering if there was a way or if I could be pointed to related research/articles etc. that directly evaluates infinite sums by evaluating a related integral. To clarify I am thinking of something along these lines: $$\sum_{i=1}^n G(i) =\int_a^b P(x)dx$$ In this scenario the infinite sum of G(i) is being evaluated by converting the problem into an equal problem of evaluating an integral which should be easier than evaluating the infinite sum in many cases. Again I am not really looking for any "tricks" for evaluating infinite sums that involve integrals I am looking for stuff that evaluates infinite sums by converting them into some type of solvable definite integral. This process would be something along the lines of how you could take the Riemann sum of something like $x^2$ and using the formula for sum of the $x^2$ function and taking a limit as n goes to infinity you could evaluate the integral of $x^2$ by evaluating a related infinite sum. Except the process I am looking for would be the reverse of this process. I would also like to mention that my question is very similar to "Is it possible to write a sum as an integral to solve it?" but all the responses to that question focused on the last part of his/her question where he/she asked for general tricks. I am solely interested in the first part of that question so I don't think my question is a repeat. Thankyou I know my question is very long but I am not always the best at explaining things and I wanted to make sure my question made sense.
1 Answers
A cool trick is to algebraiclly manipulate a sum to look like a Reimann sum.
For example : We can look at the sum $\sum_{k=0}^n \frac k {n^2}$ as a Reimann sum, on the interval $[0,1]$,with a partition that equallity "cuts" the interval into $n$ partitions, and a choice of points $t_k$ to be the right edge of each partition:
$\sum_{k=0}^n \triangle x_i f(t_k) = \sum_{k=0}^n \frac 1 n f(t_k)$, our sum fits the function $f(x)=x$, so we get : $\sum_{k=0}^n \frac 1 nt_k$ , and since $t_k$ are the right edge of each partition , we get $t_k=\frac k n$. Overall, we get $\lim_{n\to\infty} \sum_{k=0}^n \frac 1 n\frac k n=\sum_{k=0}^n \triangle x_i f(t_k) =\int_0^1x dx= \frac 1 2$
Edit : I`m not sure if this answers the question, since it was unclear if you wanted to evaluate integrals using sums or evaluate sums using integrals.
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