calculus ) Decide whether thegivene)とseries area) Žcenvergent ordivergentna1*)こ。n=2c)と8) E (- land) E(-2)2nn=1

Name: calculus ) Decide whether thegivene)とseries area) Žcenvergent ordiverg
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Description: calculus ) Decide whether thegivene)とseries area) Žcenvergent ordivergentna1*)こ。n=2c)と8) E (- land) E(-2)2nn=1

We have to decide whether the given series are convergent:a) ∑n=1∞n!n2enUsing Ratio Test, we get,…

Answered: ) Decide whether the given e) Ž series…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter3: Triangles

Section3.5: Inequalities In A Triangles

Problem 35E: Prove by the indirect method: Given: MPN is not isosceles Prove: PMPN

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calculus

) Decide whether the
given
e)と
series are
a) Ž
cenvergent or
divergent
na1
*)こ。
n=2
c)と
8) E (- lan
d) E
(-2)2n
n=1

Expert Solution

Step 1

We have to decide whether the given series are convergent:

a) $\sum_{n = 1}^{\infty} \frac{n!}{n^{2} e^{n}}$
Using Ratio Test, we get,
$\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = \lim_{n \to \infty} |\frac{(n + 1)!}{{(n + 1)}^{2} e^{n + 1}} . \frac{n^{2} e^{n}}{n!}| = \lim_{n \to \infty} \frac{n^{2}}{(n + 1) e} = \infty$
Now, as the value of the limit is greater than 1, the test tells us that the series diverges.

Step 2

b) $\sum_{n = 1}^{\infty} |\sin (\frac{1}{n^{2}})|$

Consider another series, $\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} 1$ , and let $\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} |\sin (\frac{1}{n^{2}})|$ . Clearly, $\sum_{n = 1}^{\infty} b_{n}$ is convergent.
Then, we can see that $a_{n} \leq b_{n}, \forall n$ , as, $|\sin (x)| \leq 1, \forall x .$

Therefore, we have a convergent series, $\sum_{n = 1}^{\infty} b_{n}$ , such that $a_{n} \leq b_{n}, \forall n$ , and so, by comparison test, we can say that the series $\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} |\sin (\frac{1}{n^{2}})|$ is convergent.

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