Is there any procedure to follow when determining the function of a series? This seems simple but for I can't figure it out.

$$ \frac15 + \frac18 + \frac1{11} +\frac1{14} + \frac1{17}+\cdots$$

  • $\begingroup$ What do you mean by "the function of a series"? $\endgroup$ – Fly by Night Apr 7 '13 at 20:58
  • $\begingroup$ I want to write it in summation notation. I am going to test it's convergence. $\endgroup$ – Bob Shannon Apr 7 '13 at 20:59
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    $\begingroup$ What are the coefficients? $\frac{1}{5 + 3 k}$? That series doesn't converge. $\endgroup$ – vonbrand Apr 7 '13 at 21:00
  • $\begingroup$ You have a sum of five explicitly given rationals. That can be evaluated to an explicit result using grade-school fraction rules. There's nothing to speak about convergence for here. $\endgroup$ – hmakholm left over Monica Apr 7 '13 at 21:01
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    $\begingroup$ Do you know about the harmonic series? If you do, you can tell if $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\cdots$ converges or not. Compare this to the series from your question. It may be relevant to skip the first term in one of the series. (Edit: I assumed Fly by Night's edit.) $\endgroup$ – Jeppe Stig Nielsen Apr 7 '13 at 21:05

Look at the denominators: $5, 8, 11, 14, 17, \ldots$ Can you find an expression for these?

Start by looking at the differences: $8-5=3$, $11-8=3$, $14-11=3$ and $17-14=3$. To get from one term to the next, we simply add $3$; the term-to-term rule is $+3$. That means that the sequence $5,8,11,14,17,\ldots$ is like the three times table, i.e. it resembles $3n$.

However, the three times table goes $3,6,9,12,15,\ldots$ Our sequence, $5,8,11,14,17,\ldots$, is always two bigger than the three times table. Hence the $n$-th term rule is $3n+2$.

Let's check: $3\times 1 + 2 = 5$, $3\times 2 + 2 = 8$, $3\times 3 + 2 = 11$, $3 \times 4 + 2=14$ and $3\times 5 + 2 = 17$.

If the sequence $5,8,11,14,17,\ldots$ is given by $3n+2$ then the sequence $\frac{1}{5},\frac{1}{8},\frac{1}{11},\frac{1}{14},\frac{1}{17},\ldots$ is given by:


The partials sums are then given by

$$\sum_{n=1}^p \frac{1}{3n+2}$$

Sadly, as $p \to \infty$, we don't get a sensible answer. Notice that $5n \ge 3n+2$ for all $n \ge 1$ and so

$$\frac{1}{5n} \le \frac{1}{3n+2}$$

for all $n \ge 1$. If we can show that $\sum_{n \ge 0} 1/5n$ diverges then clearly our series will diverge. Well:

$$\sum_{n \ge 1} \frac{1}{5n} = \frac{1}{5}\sum_{n \ge 1} \frac{1}{n}$$

The sum $\sum_{n \ge 1} 1/n$ is called the Harmonic Series and is well-known to diverge. Since each and every term of $\sum_{n \ge 1}1/5n$ is less than or equal to the corresponding term in the series $\sum_{n \ge 1}1/(3n+2)$, and the series $\sum_{n \ge 1}1/5n$ diverges, it follows that the series $\sum_{n \ge 1}1/(3n+2)$ also diverges.


Try to find a regular pattern. In this case it appears to be:

$f_n = \frac{1}{2+3n}, n=1,2,3,...$


Generally speaking, when I see a series like this, I try to construct a generating function of some sort. In this case:

$$f(x) = \sum_{k=1}^{\infty} \frac{x^{3 k+2}}{3 k+2}$$

You can easily show that

$$f'(x) = \frac{x^4}{1-x^3}$$

So we integrate and use $f(0)=0$ to evaluate $f(1)$. In this case, it should be clear that this value will be infinite, but it may be instructive to see how the sum diverges.

We evaluate the integral by partial fractions:

$$\frac{x^4}{1-x^3} = x \left ( \frac{1}{1-x^3} - 1\right ) = \frac{1}{3} \left ( \frac{1}{1-x} + \frac{x+1/2}{(x+1/2)^2 + 3/4} - \frac{3/2}{(x+1/2)^2 + 3/4}\right)-x$$

Doing the integration and applying the initial condition reveals

$$f(x) = \frac{1}{6} \log{(1+x+x^2)} - \frac{1}{\sqrt{3}} \arctan{\left[\frac{2 x+1}{\sqrt{3}}\right]} - \frac{1}{3} \log{(1-x)} - \frac{1}{2} x^2 + \frac{\pi}{6 \sqrt{3}}$$

So that, as $x \rightarrow 1^-$, $f(x)$ diverges logarithmically as one may expect.


There is a general procedure to compute this kind of series as detailed in Abramowitz and Stegun. The result may be expressed in terms of $\psi$ digamma functions (and their polygamma derivatives).

For a finite sum of $n$ terms you would get : $$S(n)=\frac 13\sum_{k=1}^n \frac 1{k+2/3}=\frac 13\left(\psi\left(n+1+\frac 23\right)-\psi\left(1+\frac 23\right)\right)$$ (note that this is a mere rewriting of the initial sum even if it may be useful to get numerical approximations)

Since $\psi$ is increasing on $\mathbb{R}^+$ we have $\ \psi(n+1)<\psi(n+1+2/3)<\psi(n+2)\ $ for $n\ge 0\,$ and since $\ \psi(n+1)=H_n-\gamma\ $ (with $H_n$ the $n$-th harmonic number) grows like $\,\ln(n+1)=\int \frac{dn}{n+1}\,$ $S(n)$ will go to $+\infty$ as $\,n\to +\infty$ : $$S(\infty)=\lim_{n->\infty}\frac 13\sum_{k=1}^n \frac 1{k+2/3}=\frac 13\left(\lim_{n\to\infty}\psi\left(n+1+\frac 23\right)-\psi\left(1+\frac 23\right)\right)=+\infty$$ To show a convergent result let's reproduce example $9$ of page $265$ from A&S :

\begin{align} \sum_{n=1}^\infty \frac 1{(n+1)(2n+1)(4n+1)}&=\sum_{n=1}^\infty\left(\frac 13\frac 1{n+1}-1\frac 1{n+1/2}+\frac 23\frac 1{n+1/4}\right)\\ &=-\frac 13\psi(1+1)+\psi\left(1+\frac 12\right)-\frac 23\psi\left(1+\frac 14\right)\\ \end{align} In fact using Gauss formula you may obtain a closed form for $\psi\left(1+\frac pq\right)$, some examples are at end of file. For higher orders polygamma Clausen functions may be required.

  • $\begingroup$ Could you please say a few words about why $\psi(n+5/3)-\psi(5/3) \to \infty$ as $n \to \infty$? $\endgroup$ – Fly by Night Apr 7 '13 at 22:15
  • $\begingroup$ @FlybyNight: I edited my answer with more details for the finite and infinite case. $\endgroup$ – Raymond Manzoni Apr 7 '13 at 23:42

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