3
$\begingroup$

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.

  • 1
    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.2012-01-24

2 Answers 2

10

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

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

--

$^1$ 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 $\frac{1}{2k}$.

  • 0
    @Curious: Thanks!2012-01-23
0

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.