Convergence of a Taylor series

 If a Taylor series for a function f converges on some interval, how can we determine if that series converges to the function f?Let's remind that a series converges to a particular value if the sequence of its partial sums converges to that value. In order words, the limit of the general term of the partial sum is equal to that value. For a Taylor series of a function f at a, the nth partial sum is given by pn. To determine if the Taylor series converges to f, we need to demonstrate whether

Since the remainder Rn(x) = f(x)-pn(x), the Taylor series converges to f if and only if

Convergence of a Taylor series

To prove that a Taylor series for a function f at a converges to f, we need to prove Rn(x)→0. To do this we use the bound:

Example

Solution

a. The Mclaurin series for e^x is given by 

Let's use the ratio test to determine its convergence:

Let's take the limit:

Therefore the series converges absolutely for all x, and thus, the interval of convergence is (-∞, +∞). 

Let's use the formula for the remainder of a Taylor series:

Since a =0 (Mclaurin series), we have:

❘Rₙ❘≤(M/(n + 1)!❘)x❘ⁿ⁺¹

The exponential function is an increasing function. For any I = [-d.d], the maximum value for M is e^d or M = e^d. 

Substituting M, we have:

Rₙ❘≤ e^d/(n + 1)!❘)x❘ⁿ⁺¹

Taking the limit of both sides we have:

b. Let's prove that the Mclaurin series for sinx converges to sinx. 

The Mclaurin series for sinx is given by:

Let's prove that this series is convergent and determine its convergence.

Let's take the limit of both sides, we have:

 

Let's use the formula for the  Taylor series with remainder:

Since a = 0, we have:  ❘Rₙ❘≤(M/(n + 1)!❘)x❘ⁿ⁺¹. Let's find the maximum value for M. It represents the maximum value of the absolute value des n+1 derivatives of sinx. We have:

For any x or in any interval I = [-d, d], the maximum value for M is 1. Therefore we have:

Practice