How do you use the Ratio Test on the series #sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1))# ?

Question

Emilia Aguilar · Accepted Answer

By Ratio Test, the posted series converges absolutely.

By Ratio Test:

#lim_{n to infty}|a_{n+1}/a_n|=lim_{n to infty}|{(-10)^{n+1}}/{4^{2n+3}(n+2)}cdot{4^{2n+1}(n+1)}/{(-10)^n}|#

By canceling out common factors:

#=lim_{n to infty}|{-10(n+1)}/{4^2(n+2)}|#

since #|{-10}/4^2|=5/8#, we have:

#=5/8lim_{n to infty}(n+1)/(n+2)#

by dividing the numerator and the denominator by #n#,

#=5/8 lim_{n to infty}{1+1/n}/{1+2/n}=5/8cdot 1=5/8<1#

Hence, #sum_{n=1}^{infty}{(-10)^n}/{4^{2n+1}(n+1)}# is absolutely convergent.

Isabella Ackerman · Answer

undefined To use the Ratio Test on the series $ \sum_{n=1}^\infty \frac{(-10)^n}{4^{2n+1}(n+1)} $, we evaluate the limit:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

where $ a_n $ represents the $ n $-th term of the series.

In this case, $ a_n = \frac{(-10)^n}{4^{2n+1}(n+1)} $.

Now, let's find $ a_{n+1} $:

$$ a_{n+1} = \frac{(-10)^{n+1}}{4^{2(n+1)+1}((n+1)+1)} = \frac{(-10)^{n+1}}{4^{2n+3}(n+2)} $$

Now, let's find the ratio $ \frac{a_{n+1}}{a_n} $:

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(-10)^{n+1}}{4^{2n+3}(n+2)}}{\frac{(-10)^n}{4^{2n+1}(n+1)}} $$

$$ = \frac{(-10)^{n+1}}{4^{2n+3}(n+2)} \cdot \frac{4^{2n+1}(n+1)}{(-10)^n} $$

$$ = \frac{(-10)(n+1)}{4^2(n+2)} $$

Now, let's take the limit as $ n $ approaches infinity:

$$ \lim_{n \to \infty} \frac{(-10)(n+1)}{4^2(n+2)} = \lim_{n \to \infty} \frac{-10n - 10}{16n + 32} = \frac{-10}{16} = -\frac{5}{8} $$

Since the absolute value of this ratio is less than 1, specifically $ \frac{5}{8} $, by the Ratio Test, the given series $ \sum_{n=1}^\infty \frac{(-10)^n}{4^{2n+1}(n+1)} $ converges.