# Convergence of $\sum_{n=1}^\infty\frac{1}{2\cdot n}$

It is possible to deduce the value of the following (in my opinion) converging infinite series? If yes, then what is it?

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

where n is an integer. Sorry if the notation is a bit off, I hope youse get the idea.

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The series diverges: it’s simply $1/2$ of the harmonic series, which is a standard example of a divergent series. – Brian M. Scott Jan 23 '12 at 11:38
Opinions aren't worth much in mathematics - do you have any reason for thinking that series converges? – Gerry Myerson Jan 23 '12 at 11:44
OK, but usually when people have opinions, those opinions are based on something. I'm "curious" as to the basis for your opinion that the series converges. – Gerry Myerson Jan 23 '12 at 12:00
@GerryMyerson - and surely the basis of an opinion must be based on something else etc etc. But seriously.. I got a decent answer and see no benefit in arguing over supposed merits or shortcomings of reasoning that lead to this question. Thanks again, and take care. – Curious Jan 23 '12 at 13:14
I'm glad you got a helpful answer. The benefit in discussing the reasoning that led to a question is that you might learn something about how to decide mathematical questions that interest you, or how to phrase such questions when asking them of other people. I don't see the benefit in shutting down discussion of unresolved points. – Gerry Myerson Jan 24 '12 at 1:26

The series is not convergent, since it is half of the harmonic series which is known to be divergent$^1$.

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

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$^1$ The sum of the following $k$ terms is greater or equal to $\frac{1}{2}$

$$\frac{1}{k+1}+\frac{1}{k+2}+\ldots +\frac{1}{2k-1}+\frac{1}{2k}\geq k\times \frac{1}{2k}=\frac{1}{2},$$

because each term is greater or equal to $\frac{1}{2k}$.

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I'd upvote this if I could but I don't have the necessary reputation just yet. – Curious Jan 23 '12 at 11:54
@Curious: Thanks! – Américo Tavares Jan 23 '12 at 15:11

To me, the easiest way to see that the harmonic series diverges is to use the Integral test. Then you do not have to deal with coming up with a formula.

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