How do you test the alternating series #Sigma (-1)^nsqrtn/(n+1)# from n is #[1,oo)# for convergence?

Question

Elijah Abram · Accepted Answer

The series:

#sum_(n=1)^oo (-1)^n sqrt(n)/(n+1)#

is convergent

The series:

#sum_(n=1)^oo (-1)^n sqrt(n)/(n+1)#

is convergent

The series:

#sum_(n=1)^oo (-1)^n sqrt(n)/(n+1)#

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

#sum_(n=1)^oo (-1)^n a_n#

is convergent if:

(i) #lim_(n->oo) a_n = 0#

(ii) #a_(n+1) <= a_n#

in our case:

#lim_(n->oo) sqrt(n)/(n+1) = lim_(n->oo)1/((n+1)/ sqrt(n)) = lim_(n->oo)1/((sqrt(n) +1/ sqrt(n))) = 0#

so the first condition is satisfied. For the second we analyze the function:

#f(x) = sqrt(x)/(x+1)#

and calculate the derivative:

#(df)/(dx) = ((x+1)/(2sqrt(x)) - sqrt(x))/(x+1)^2= (x+1-2x)/(2sqrt(x)(x+1)^2)= - (x-1)/(2sqrt(x)(x+1)^2)#

we can see that #(df)/(dx) < 0# for #x in (1,+oo)# therefore the function is strictly decreasing in that interval and we have:

#f(n+1) < f(n)#

that is:

#sqrt(n+1)/(n+2) <= sqrt(n)/(n+1)#

and also the second condition is satisfied.

Elijah Abram · Answer

undefined To test the convergence of the alternating series $ \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1} $, we can use the Alternating Series Test, which states that if the series alternates in sign and the terms decrease in absolute value, then the series converges.

We first need to check that the terms $ \frac{\sqrt{n}}{n+1} $ decrease as $ n $ increases. We can do this by examining the ratio of consecutive terms:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\sqrt{n+1}}{n+2} \times \frac{n+1}{\sqrt{n}} $$

Simplifying this expression and taking the limit will give us a result of 1, which means the terms do not strictly decrease. Therefore, we cannot conclude convergence based on the Alternating Series Test alone.

Instead, we can use the Limit Comparison Test. Let's consider the series $ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $, which is a p-series with $ p = \frac{1}{2} $. This series diverges.

We'll take the limit of the ratio of the terms of the given series to the terms of the divergent series:

$$ \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n+1}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{\sqrt{n} \cdot \sqrt{n}}{(n+1)} = 1 $$

Since the limit is a finite, positive number, by the Limit Comparison Test, our original series $ \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1} $ has the same convergence behavior as the divergent series $ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $, which means it also diverges.