For what values of $x$ is the series convergent $\sum_{n=1}^∞ \frac{(-1)^n(x+3)^n}{n \cdot 5^n}$

Question

$$\sum_{n=1}^∞ \frac{(-1)^n(x+3)^n}{n \cdot5^n}$$ Can you help me find the values of $x$ for which the series converges.

It feels like it can be done by root method which I tried but what should I do of the $n$ in the denominator?

Answer 1

$$\frac{(-1)^n(x+3)^n}{n.5^n}=\dfrac{\left(-\dfrac{x+3}5\right)^n}n$$

Now $$\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$$ converges if $$-1\le y<1$$

See

What is the correct radius of convergence for $\ln(1+x)$?

or Taylor series for $\log(1+x)$ and its convergence

Answer 2

Let $a_n=\frac{(-1)^n}{n*5^n}$. Then $|a_n|^{1/n} \to 1/5$. Hence the power series has radius of convergence $=5$.

The power series converges absolutely for $|$x$+$3$|$<$5$ and diverges for $|x+3$|$>5$.

It is your turn to investigate the cases $x+3=\pm 5$.

Answer 3

$$|\frac{x_{n+1}}{x_n}|=|\frac{\frac{(-1)^{n+1}(x+3)^{n+1}}{(n+1)\cdot5^{n+1}}}{\frac{(-1)^{n}(x+3)^{n}}{n\cdot5^n}}|= |\frac{n(x+3)}{5(n+1)}|$$

Taking the limit of this as $n$ tends to infinity gives that (ratio test) the sum converges for $|x+3| < 5$

You also know that (again, by ratio test), that if $|x+3| > 5$ then the sum will diverge;

What is left is to check the case $|x+3| = 5$ i.e $(x+3) = -5$ or $(x+3) = 5$ which result in the alternate harmonic series and the harmonic series respectively. The first can be shown to be convergent by the Alternating Series Test, while the second is divergent (a standard result)

Answer 4

HINT: solve it by Power series $ \sum a_n (x-x_o)^n$ , where $ a_n$=$\frac{(-1)^n}{n.5^n}$ and $x_o=-3$

Answer 5

For power series,$$\sum_{n=1}^{\infty}a_nX^n$$

$a_n=\frac {(-1)^n}{n.5^n}$ , and $X=-(x+3)$

By ratio test, $$\lim_{n\to∞}|{\frac{a_{n+1}}{a_n}}|=\frac 15$$ Radius of convergence $=5$

So the series is convergent for $|x+3|∈(-5,5)$ ie. $x∈(-8,2)$

Checking on $x=-8$, the series sum $=\sum_{n=1}^{\infty}\frac 1n$ which diverges.

at $x=2$, series sum $=\sum_{n=1}^{\infty}\frac {(-1)^n}n$ which converges.

Hence answer is $x∈(-8,2]$