Does (-2)^n/( 3^n+1) converge or diverge?

1 answer:

3 0

$\left|\dfrac{(-2)^n}{3^n+1}\right|=\dfrac{2^n}{3^n+1}$

As $n\to\infty$ , the sequence $a_n=\left(\dfrac23\right)^n$ converges to zero.

If you're talking about the infinite series

$\displaystyle\sum_{n\ge0}\dfrac{(-2)^n}{3^n+1}$

well we've shown by comparison that this series must also converge because we know any geometric series $\sum\limits_n r^n$ will converge as long as $|r|$ .

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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

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Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$