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We can use the ratio test for series to find the values for which the series converges.

We have that a series

converges if

It means that a series converges if its terms grow slower than the terms of a convergent geometric series (at least after several terms). If the limit is larger than one the terms get bigger and bigger, and the series will diverge, and if the limit is 1, then the test is inconclusive.

If we for example look at

So

and we thus have that

This should be less than 1 for the series to converge, and we thus have that the series converges when

The radius of convergence is thus 1.

We must check the endpoints of the interval separately. In this case, it is easy to see that neither f(1) nor f(-1) will converge, and the interval of convergence will this be (–1,1) (or ]–1,1[ or {x| –1<x<1}, depending on the way of writing it).

**Logarithms**

For

we have

and thus

We yet again get that the series converges when

So the radius of convergence is thus 1.

For the endpoints we have

that indeed converges to ln(2). But for the other endpoint we have ln(1-1)=ln(0) that is undefined, and we also have that

converges. So the interval of convergence is (1,–1] or ]1,1].

**The exponential function**

We have

So

and

This is less than one, so the series will converge for any value of *x*. It will thus have an infinite radius of convergence. As you can verify this is true for the sine and cos functions too.

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