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Let $ f(x) = \sum_{n = 1}^{\infty} \frac {x^n}{n^2} $Find the intervals of convergence for $ f, f', $ and $ f''. $
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Calculus 2 / BC
Chapter 11
Infinite Sequences and Series
Section 9
Representations of Functions as Power Series
Sequences
Series
Joshua G.
May 21, 2021
Wouldn't f"(1) converge since it = lim n>>inf (1^(n-2)*(n-1))/n = lim n>>inf (n-1)/n = 0 ?
Campbell University
Harvey Mudd College
Baylor University
Idaho State University
Lectures
01:59
In mathematics, a series is, informally speaking, the sum of the terms of an infinite sequence. The sum of a finite sequence of real numbers is called a finite series. The sum of an infinite sequence of real numbers may or may not have a well-defined sum, and may or may not be equal to the limit of the sequence, if it exists. The study of the sums of infinite sequences is a major area in mathematics known as analysis.
02:28
In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. Formally, a sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequences) or the set of the first "n" natural numbers (for a finite sequence). A sequence can be thought of as a list of elements with a particular order. Sequences are useful in a number of mathematical disciplines for studying functions, spaces, and other mathematical structures using the convergence properties of sequences. In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers.
01:07
$$f(x)=\sum_{n=0}^{\infty}…
11:16
Let $f(x)=\sum_{n=0}^{\inf…
02:42
02:04
$$\quad \text { Let } f(x)…
02:16
Find the intervals of conv…
09:36
Letf(x) = n2 ` n = 1
The problem is that f of x is equal to some and from 1 to infinity x, to n over n squared finding the intervals of convergence of f, f, prime and f prim prim. So, first we need to find the the raiders of convergence so for this function we use x, plus 1, over n plus 1 square over x, o n over n squared is absolute limit and goes to infinity. This is equal to absolve of x, 2 n plus 1 over n plus 1 square times square over x, 2 n absalom limit and goes to infinity. Here we can cancel out x to nets. The limit is equal to absolute value of x by ratio test. If this is less than 1 to function, f of x is converted to absolutely absolutely point, so the radius of convergence is equal to 1. Then we we check and points negative 1 and 1 for the function of x on x is equal to 1 negative 1. Half of x is equal to some from 1 to infinity. Negative 1 to n over n squared 1 x is equal to 1 of x is equal to sam and from 1 to infinity 1 over n square. Since here the power of n is 2, which is greater than 1 on, so alphax is converted absolutely to both of them murders so for flax. Interval of convergence is bracket negative, 1 and 1. So as from half the value of convergence, s also 1, and then we check its end points on x, so f of x from is equal to some and from 1 to infinity. This is x, 2 minus 1. Over n, i 1 is equal to some and from 1 to infinity negative 1 to the power of minus 1 on the balais test, this series converges of from at a point of 1. This is equal to some 1 over n, since the power of n is equal to 1, which is less than 1 divergent, so for f, x, prim interval of convergence is bracket. Negative 1 and 1 parentheses for sack derivative in the radius of convergence is also 1, but we also need to check its end points. The first from from is equal to some and from 2 to infinity x, to m minus 2 over n times and minus y. So we can see from prime at this point. Negative 1 is equal to some negative 1 times n minus 1 net 1 to the power of minus 2 times 1 over, since the limit of the general term is not equal to 0. The limit of this 1 is not equal to 0 point, so this series des for the time reason we can see from primate .1 also diverges so for the function of x, from from interval of convergence is parenthesis negative 1 and 1.
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