Formula for partial sum of (10)/(10n+1) 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?
 A: 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 terms 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.
