Find the interval of convergence of the power series. ∑n=1∞(−1)nxnn

The given series is ∑n=1∞(−1)nxnn Compare the above series with its standard form ∑n=1∞(−1)nanxn and obtain that an=1n Obtain the radius of convergence as follows. R=limn→∞anan+1 =limn→∞1n×n+11 =limn→∞(1+1n) =1+0 =1 Thus, the radius of convergence is 1. That is, the series ∑n=1∞(−1)nxnn converges for all x in (-1,1). Now check at its end points as follows. at x=1, the series become ∑n=1∞(−1)nn which is convergent. at x=-1, the series become ∑n=1∞(−1)2nn=∑n=1∞1n which is divergent. Thus, the interval of convergence is (−1, 1].

Best solutions to - "Find the interval of convergence of the power..."

Amari Flowers

Answered question

2020-10-27

Find the interval of convergence of the power series.

\sum_{n = 1}^{\infty} \frac{(- 1)^{n} x^{n}}{n}

Answer & Explanation

aprovard

Skilled2020-10-28Added 94 answers

The given series is $\sum_{n = 1}^{\infty} \frac{(- 1)^{n} x^{n}}{n}$
Compare the above series with its standard form $\sum_{n = 1}^{\infty} (- 1)^{n} a_{n} x^{n}$ and obtain that $a_{n} = \frac{1}{n}$
Obtain the radius of convergence as follows.
$R = lim_{n \to \infty} \frac{a_{n}}{a_{n + 1}}$
$= lim_{n \to \infty} \frac{1}{n} \times \frac{n + 1}{1}$
$= lim_{n \to \infty} (1 + \frac{1}{n})$
$= 1 + 0$
$= 1$
Thus, the radius of convergence is 1.
That is, the series $\sum_{n = 1}^{\infty} \frac{(- 1)^{n} x^{n}}{n}$ converges for all x in (-1,1).
Now check at its end points as follows.
at x=1, the series become $\sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{n}$ which is convergent.
at x=-1, the series become $\sum_{n = 1}^{\infty} \frac{(- 1)^{2 n}}{n} = \sum_{n = 1}^{\infty} \frac{1}{n}$ which is divergent.
Thus, the interval of convergence is (−1, 1].

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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