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I was wondering, for example,

Can:

$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$

Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.

But actually writing an integral form. Like

$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{a}^{b} g(x) \space dx$$

What are some general tricks in finding infinite sum series.

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  • $\begingroup$ math.stackexchange.com/questions/1002440/… $\endgroup$ – lab bhattacharjee Nov 3 '14 at 13:26
  • $\begingroup$ @labbhattacharjee, I did not meant that. I know the solution to this, I was just asking if in general it is possible to write a sum as an actual integral. $\endgroup$ – Amad27 Nov 3 '14 at 13:28
  • $\begingroup$ You can trivially write the sum as an integral using the Iverson bracket (add a factor of $[n \in \mathbb{N}]$ to the integrand). This ignores the question of how to evaluate the resulting integral, of course. $\endgroup$ – chepner Nov 3 '14 at 19:10
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    $\begingroup$ "I am NOT talking about a method for using tricks with integrals." "But actually writing an integral form." "What are some general tricks" Combining these quotes with the accepted answer that does not seem to be a general trick, I'm a bit confused on what this question is asking. $\endgroup$ – JiK Nov 4 '14 at 8:28
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    $\begingroup$ @Amad27 $\int_\mathbb{N}\frac{d \mu}{(3n-1)(3n+2)}$ where $\mu$ is the counting measure on $\mathbb{N}$. It doesn't give you anything you didn't already have though. I didn't really mean it seriously although it is true. $\endgroup$ – Tim Seguine Nov 5 '14 at 17:21
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A General Trick

A General Trick for summing this series is to use Telescoping Series: $$ \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\frac13\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{3n-1}-\frac1{3n+2}\right)\\ &=\frac13\lim_{N\to\infty}\left[\sum_{n=1}^N\frac1{3n-1}-\sum_{n=1}^N\frac1{3n+2}\right]\\ &=\frac13\lim_{N\to\infty}\left[\sum_{n=0}^{N-1}\frac1{3n+2}-\sum_{n=1}^N\frac1{3n+2}\right]\\ &=\frac13\lim_{N\to\infty}\left[\frac12-\frac1{3N+2}\right]\\ &=\frac16 \end{align} $$


An Integral Trick

Since $$ \int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n $$ for $n\gt0$, we can write $$ \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\ &=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\ &=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\ &=\frac16 \end{align} $$

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    $\begingroup$ I think this is a better "trick" for dealing with sums. Integral "tricks" are nice however integrals and infinite series' are very different in what they calculate and manipulating a sum or integral on its own without switching is preferred. $\endgroup$ – Ali Caglayan Nov 3 '14 at 16:23
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    $\begingroup$ @Alizter: For the most part, I agree. However, sometimes pure series manipulation can be extremely complicated, and the proper integral representation of a sum can be useful. However, in this case, I think staying with series manipulation is easiest. That being said, I have added an integral approach, as well. $\endgroup$ – robjohn Nov 3 '14 at 18:40
  • $\begingroup$ The sum under the first integral could have been computed as a telescoping series either. Considering this, I think the use of integrals in the second solution is completely void. Edit: I mean exactly what Henning Makholm points out under the other answer. $\endgroup$ – Adayah Nov 4 '14 at 19:41
  • $\begingroup$ @Adayah: My reply to Henning was meant as an agreement. I first posted only the telescoping series, but then added an integral approach to satisfy the first part of the question. In any approach where one breaks up the summand using partial fractions, it could be said that, at that point, the answer could be computed as a telescoping sum. $\endgroup$ – robjohn Nov 4 '14 at 20:48
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Since $\int_{0}^{1}x^k\,dx = \frac{1}{k+1}$, $$\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)=\frac{1}{3}\int_{0}^{1}x^{3n-2}(1-x^3)\,dx,$$ so, summing over $n$: $$\sum_{n=1}^{+\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\int_{0}^{1}x\,dx=\frac{1}{6}.$$

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  • $\begingroup$ I thought we need uniform convergence in order to interchange the limit and integral. The power series is uniformly convergent inside the radius of convergence, how to pass it to the whole interval $[0,1]$? $\endgroup$ – John Nov 3 '14 at 13:45
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    $\begingroup$ @JohnZHANG Actually no, Fubini and Tonelli's theorems allow this for a monotone sequence supposedly, I believe. $\endgroup$ – Amad27 Nov 3 '14 at 13:45
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    $\begingroup$ Nice trick for the given sum, but this still doesn't answer the bold-marked question of general tricks. $\endgroup$ – Ruslan Nov 3 '14 at 16:34
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    $\begingroup$ Isn't the integral just a detour here? The operative step is exactly the same telescoping that could have been done without rewriting the terms into integrals. $\endgroup$ – Henning Makholm Nov 4 '14 at 10:11
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    $\begingroup$ @HenningMakholm: smoke and mirrors. $\endgroup$ – robjohn Nov 4 '14 at 12:07
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Actually writing it as an integral, as asked for:

$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{1}^{\infty} \frac{1}{(3\lfloor x\rfloor-1)(3\lfloor x\rfloor+2)} dx$$

This probably won't help with finding the value, though.

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  • $\begingroup$ why won't it help finding the value? $\endgroup$ – Amad27 Nov 4 '14 at 13:20
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    $\begingroup$ @Amad27: I don't see a way it would. If you can find one, then more power to you, I suppose ... $\endgroup$ – Henning Makholm Nov 4 '14 at 13:33
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    $\begingroup$ @Amad27 Methods for solving integrals are poorly suited for integrating functions that are non-continuous. The usual approach for integrating functions like the one here is to separately integrate over each interval where it is continuous. Which brings us back to the sum form. $\endgroup$ – Rafał Dowgird Nov 4 '14 at 15:11
  • $\begingroup$ @Amad27 It is quite literally equivalent to the original sum in a trivially useless manner XD $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 2:11
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In such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.
Given that $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\sum_{n=1}^{\infty} T_{n}$$ Where, $T_{n}$ is the $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follows $$T_{n}=\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)$$ Now, we have $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right) $$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -\frac{1}{\infty}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2}\right]=\color{blue}{\frac{1}{6}}$$

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We can indeed write the sum as an integral, after research. Consider:

Find: $\psi(1/2)$

By definition:

$$\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$$

The required $z$ is $z = -\frac{1}{2}$

so let $z = -\frac{1}{2}$

$$\psi(1/2) = -\gamma + \sum_{n=1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$

Simplify this: $$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{n(2n - 1)}$$

The sum seems difficult, but really isnt.

We can telescope or:

$$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$

Let $x \rightarrow x^2$

$$\frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$

Integrate once:

$$\tanh^{-1}(x) = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$$

Integrate again:

$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = 2\int \tanh^{-1}(x) dx$$

From the tables, the integral of $\tanh^{-1}(x)$

$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = \log(1 - x^2) + 2x\tanh^{-1}(x)$$

Take the limit as $x \to 1$

$$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)} = \log(4)$$

$$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)}$$

$$\psi(\frac{1}{2}) = -\gamma - \log(4)$$

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  • $\begingroup$ I am the OP per say. This is a general trick. I converted the sum into an integral. Please read carefully. $\endgroup$ – Amad27 Dec 21 '14 at 8:15
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Yes you can use the Euler Maclaruin formula to write the sum as an integral plus an infinite number of derivatives. I remember deriving this for my self when I was younger and being very pleased with myself.

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This particular sum could be solved because you had two terms $ax+b$ and $ax+c$ and the difference between c and b is equal to a (I think it would work in a slightly more complicated way if it was a not-too-large multiple of a).

If you want numerical values in general cases, and the sum doesn't converge quickly for your taste, or you want just a partial sum, you can use that

$$\displaystyle f (k) = \int_{k-1/2}^{k+1/2} f(k) dx ≈ \int_{k-1/2}^{k+1/2} f(x) dx$$

and therefore

$$\displaystyle \sum_{k=n}^{m} f(k) ≈ \int_{n-1/2}^{m+1/2} f(x) dx$$

Assuming that you can solve the integral in closed form, if you let

$$\displaystyle g (k) = f(k) - \int_{k-1/2}^{k+1/2} f(x) dx$$

then

$$\displaystyle \sum_{k=n}^{m} f(k) = \int_{n-1/2}^{m+1/2} f(x) dx + \sum_{k=n}^{m} g(k)$$

$g (k)$ will usually converge much faster than $f (k)$.

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