How to convert the limit of a series into an integral?

[Adesh]

If I have a limit of a series then how can I convert it into integral. I know to convert a sum into an integral there must be Δx multiplied to each term and this must go zero. Can you please explain me the conversion of limit of series (normal series with no Δx) into an integral.
Thank you.

 

[fresh_42]

If I have a limit of a series then how can I convert it into integral. I know to convert a sum into an integral there must be Δx multiplied to each term and this must go zero. Can you please explain me the conversion of limit of series (normal series with no Δx) into an integral.
Thank you.

Are you looking for something like https://en.wikipedia.org/wiki/Integral_test_for_convergence ?

 

[Adesh]

Are you looking for something like https://en.wikipedia.org/wiki/Integral_test_for_convergence ?

I’m looking for solving the series. Thank you

 

[Stephen Tashi]

I’m looking for solving the series.

Are you asking about "summing the series" - or are you asking about "testing the convergence of the series"?

To test the convergence of a series using the integral test, we do not need to sum it.

 

[Adesh]

Are you asking about "summing the series" - or are you asking about "testing the convergence of the series"?

To test the convergence of a series using the integral test, we do not need to sum it.

I’m asking how to take the limit of a sum of a convergent series. Sir I’m sorry for ambiguity but I’m trying to be clear as much as possible.

 

[fresh_42]

I’m asking how to take the limit of a sum of a convergent series. Sir I’m sorry for ambiguity but I’m trying to be clear as much as possible.

There is no general formula $\sum_{\mathbb{N}} f(n) = \int_a^b g(x)\,dx\,.$

 

[Adesh]

There is no general formula $\sum_{\mathbb{N}} f(n) = \int_a^b g(x)\,dx\,.$

Can we do this? Where is dx in summation, sir

 

[fresh_42]

Can we do this? Where is dx in summation, sir

That's the problem, in the sum we have $dx=1$ which is not the same as what the limit $dx$ actually is. Thus the integral criterion and I'm sure there are some error estimations for it, too, is as best as we can get, as far as I know. If there was such a formula, then it certainly would be part of all curricula.

 

[Stephen Tashi]

I’m asking how to take the limit of a sum of a convergent series.

See if you like the Euler-Mclaurin summation formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula

 

[Adesh]

See if you like the Euler-Mclaurin summation formula https://en.wikipedia.org/wiki/Euler–Maclaurin_formula

I was solving this problem
lim. ∑1/(n+k). {k varies from zero to m). m= nb
n -> ∞

What most people did in my university or even on internet is that they wrote it like this ∫1/(n+x) dx (limit zero to m) m= nb
(b is a constant)
I'm wondering how can you change it into integral as difference between partitions is one and is not changing. I myself used Euler-Maclaurin formula for solving this.

 

[Mark44]

I was solving this problem
lim. ∑1/(n+k). {k varies from zero to m). m= nb
n -> ∞

What most people did in my university or even on internet is that they wrote it like this ∫1/(n+x) dx (limit zero to m) m= nb
(b is a constant)
I'm wondering how can you change it into integral as difference between partitions is one and is not changing. I myself used Euler-Maclaurin formula for solving this.

They are not evaluating the summation by writing a related integral. Most likely, they are using the integral to get an upper bound (or possibly upper and lower bounds) on the summation.

 

[Adesh]

They are not evaluating the summation by writing a related integral. Most likely, they are using the integral to get an upper bound (or possibly upper and lower bounds) on the summation.

Sir can you please explain elaborately the conversion. Here is the link where he uses the conversion of series to an integral for solving the problem.

 

[Mark44]

If I have a limit of a series then how can I convert it into integral. I know to convert a sum into an integral there must be Δx multiplied to each term and this must go zero. Can you please explain me the conversion of limit of series (normal series with no Δx) into an integral.

The guy in the video is evaluating the summation by converting the summation into the Riemann sum that is equal to a definite integral. See this Wikipedia article: https://en.wikipedia.org/wiki/Riemann_sum

In the video, he starts with $\lim_{n \to \infty} \sum_{r = 0}^{n - 1} \frac 1 {\sqrt{n^2 - r^2}}$
In the next step, he writes a Riemann sum and the definite integral the Riemann sum is equal to.
$\lim_{n \to \infty} \sum_{r = 0}^{n - 1} f(\frac r n) \frac 1 n = \int_0^1 f(n) dn$
He then works to convert the expression in the summation he's working on, $\frac 1 {\sqrt{n^2 - r^2}}$, to make it look like $f(\frac r n)\frac 1 n$. That fraction $\frac 1 n$ is the piece that you were missing, and which confused the people who have responded to your question.That fraction in the summation plays the same role as $dx$ in the integral.

 

[Adesh]

The guy in the video is evaluating the summation by converting the summation into the Riemann sum that is equal to a definite integral. See this Wikipedia article: https://en.wikipedia.org/wiki/Riemann_sum

In the video, he starts with $\lim_{n \to \infty} \sum_{r = 0}^{n - 1} \frac 1 {\sqrt{n^2 - r^2}}$
In the next step, he writes a Riemann sum and the definite integral the Riemann sum is equal to.
$\lim_{n \to \infty} \sum_{r = 0}^{n - 1} f(\frac r n) \frac 1 n = \int_0^1 f(n) dn$
He then works to convert the expression in the summation he's working on, $\frac 1 {\sqrt{n^2 - r^2}}$, to make it look like $f(\frac r n)\frac 1 n$. That fraction $\frac 1 n$ is the piece that you were missing, and which confused the people who have responded to your question.That fraction in the summation plays the same role as $dx$ in the integral.

Sir, is there any other method to solve it. I shall highly appreciate it.

 

[Mark44]

Sir, is there any other method to solve it. I shall highly appreciate it.

I don't believe there is.

 

[Adesh]

I don't believe there is.

Thank you

 

[WWGD]

I was solving this problem
lim. ∑1/(n+k). {k varies from zero to m). m= nb
n -> ∞

What most people did in my university or even on internet is that they wrote it like this ∫1/(n+x) dx (limit zero to m) m= nb
(b is a constant)
I'm wondering how can you change it into integral as difference between partitions is one and is not changing. I myself used Euler-Maclaurin formula for solving this.

The guy in the video is evaluating the summation by converting the summation into the Riemann sum that is equal to a definite integral. See this Wikipedia article: https://en.wikipedia.org/wiki/Riemann_sum

In the video, he starts with $\lim_{n \to \infty} \sum_{r = 0}^{n - 1} \frac 1 {\sqrt{n^2 - r^2}}$
In the next step, he writes a Riemann sum and the definite integral the Riemann sum is equal to.
$\lim_{n \to \infty} \sum_{r = 0}^{n - 1} f(\frac r n) \frac 1 n = \int_0^1 f(n) dn$
He then works to convert the expression in the summation he's working on, $\frac 1 {\sqrt{n^2 - r^2}}$, to make it look like $f(\frac r n)\frac 1 n$. That fraction $\frac 1 n$ is the piece that you were missing, and which confused the people who have responded to your question.That fraction in the summation plays the same role as $dx$ in the integral.

With the small caveat that the dx are not required to be of uniform width ( here 1/n) , only that they all have width approaching 0 in the limit.

 

[WWGD]

Afesh, why don't you try replicating the method , dividing the denominator by n and mikick the rest of the approach in the video described by mark44 and see where that leads you?