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I'm trying to find $S_n$ of an infinite series, and I'm having trouble. Here is the equation:

$$\sum_{n=1}^\infty \frac{10}{10n+1}$$

This gives me these terms:

$$S_n = \frac{10}{11}+\frac{10}{21}+\frac{10}{31}+\frac{10}{41}+ ... + \frac{10}{10n+1}$$

After I calculate the terms of $S_n$, I get:

$$S_n = \frac{10}{11},\frac{20}{32},\frac{30}{63},\frac{40}{104}, ... $$

Obviously the top is 10n, but I'm having trouble with the bottom. I recognize a pattern in the differences of the terms, mainly that each is separated by the previous difference + 10:

$$S_2-S_1=21$$

$$S_3-S_2=31$$

$$S_4-S_3=41$$

But I have no idea how to translate that into a formula. Note that I am aware that the series diverges, but I would still like to create a formula with which I can take the limit of infinity to verify that it diverges. Any suggestions?

EDIT: Apparently I'm asking the wrong question. What I'm essentially trying to figure out is how to determine whether the series converges or diverges based on the information available. I can use intuition to come to the conclusion it's divergent, but how do I do it mathematically?

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  • $\begingroup$ the series is infinite. $\endgroup$ – mookid Nov 7 '14 at 23:47
  • $\begingroup$ $$\sum \frac{10}{10n+1}\geq \sum \frac{1}{2n}$$ then the series DOSEN'T converge. $\endgroup$ – idm Nov 7 '14 at 23:48
  • $\begingroup$ I should have specified, and I'll edit the question: I'm aware that the series is divergent, but I would like to create a formula and take the limit to infinity to confirm. $\endgroup$ – Todd Nov 7 '14 at 23:49
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The $S=\sum_{k=1}^\infty \frac{10}{10k+1}$ series is divergent. The parial sums have closed-form. Let denote

$$S_n = \sum_{k=1}^n \frac{10}{10k+1}.$$

Then in them of digamma function

$$S_n = \psi\left(n+\frac{11}{10}\right) - \psi\left(\frac{1}{10}\right) - 10.$$

Here $\psi\left(\frac{1}{10}\right)$ has an elementary closed-form, but I don't know about an alternate form of the other $\psi$ term. The first few elements in the sequence are

$$0, \frac{10}{11},\frac{320}{231},\frac{12230}{7161},\frac{573040}{293601},\frac{3573450}{1663739},\frac{234617840}{101488079},\frac{17672747430}{7205653609},\dots$$

In general if the following sum exists, then:

$$\sum_{k=1}^n \frac{a}{bk+c} = \frac ab \left( \psi\left(\frac cb + n+1\right) - \psi\left(\frac cb + 1\right) \right).$$


To answer the modified question we could use direct comparison test. As @idm mentioned, since

$$\frac{10}{10n+1}\geq\frac{1}{2n} \geq 0,$$

and because $\sum \frac{1}{2n}$ diverges it follows that $\sum \frac{10}{10n+1}$ also diverges.

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  • $\begingroup$ Yikes. That is way more advanced than I bargained for - I absolutely have no idea what is going on there. I think I'm asking the wrong question. I suppose the question I should be asking is, how do I use mathematical means (rather than intuition) to tell if that series diverges without creating a formula with which to run a limit test? $\endgroup$ – Todd Nov 8 '14 at 0:21
  • $\begingroup$ @Todd All right, I've answered your other question too. $\endgroup$ – user153012 Nov 8 '14 at 0:36
  • $\begingroup$ Thank you for your responses. That is an example of the Squeeze Theorem, correct? Also, how do we know $\sum{\frac{1}{2n}}$ diverges - is that just common knowledge that I missed, or is there a specific reason you used $\sum{\frac{1}{2n}}$ in this instance? $\endgroup$ – Todd Nov 8 '14 at 0:52

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