In the ratio test, we check to see if the series suma_k converges or diverges by examining the ratio a_(k+1)/a_k as krarroo
If lim_(krarroo)a_(k+1)/a_k > 1 then the series diverges.
If 0<= lim_(krarroo)a_(k+1)/a_k < 1 then the series converges.
If lim_(krarroo)a_(k+1)/a_k = 1 then the test does not work and a new method is needed.
In this case, a_k = ((2k)!)/k^(2k) so
lim_(krarroo)a_(k+1)/a_k = lim_(krarroo)(((2(k+1))!)/(k+1)^(2(k+1)))/(((2k)!)/k^(2k))
=>lim_(krarroo)a_(k+1)/a_k= lim_(krarroo)((2k+2)!)/((2k)!)*k^(2k)/(k+1)^(2(k+1))
((2k+2)!)/((2k)!) = (k+1)(k+2) and
k^(2k)/(k+1)^(2(k+1))=(k/(k+1))^(2k) * 1/(k+1)^2
So, multiplying, we have
lim_(krarroo)a_(k+1)/a_k= lim_(krarroo)(k+2)/(k+1)*(k/(k+1))^(2k)
lim_(krarroo)(k+2)/(k+1) = 1 and
lim_(krarroo)(k/(k+1))^(2k) = e^(-2) (see below for how to solve this part)
Then, multiplying gives us
lim_(krarroo)a_(k+1)/a_k= 1*e^-2 = e^-2
As 0 <= e^-2 < 1, by the ratio test, the series sum((2k)!)/k^(2k) converges.
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To find lim_(krarroo)(k/(k+1))^(2k) we will first look at lim_(krarroo)(k/(k+1))^k
First, let's deal with the k exponent:
lim_(krarroo)(k/(k+1))^k = lim_(krarroo)e^(ln((k/(k+1))^k)
=>lim_(krarroo)(k/(k+1))^k = lim_(krarroo)e^(kln(k/(k+1)))=e^(lim_(krarroo)kln(k/(k+1)))
Now, looking at lim_(krarroo)kln(k/(k+1)), we will deal with the natural log using L'Hopital's rule.
lim_(krarroo)kln(k/(k+1))= lim_(krarroo)ln(k/(k+1))/(1/k)
As lim_(krarroo)ln(k/(k+1)) = lim_(krarroo)1/k=0 we can apply L'Hopital's rule to obtain
lim_(krarroo)ln(k/(k+1))/(1/k) = lim_(krarroo)(((k+1)/k)*(k+1-k)/(k+1)^2)/(-1/k^2)
Simplifying, we get
lim_(krarroo)kln(k/(k+1)) = lim_(krarroo)-k/(k+1) = -1
Then we substitute back to obtain
lim_(krarroo)(k/(k+1))^k = e^-1
Thus
lim_(karroo)(k/(k+1))^(2k) = lim_(krarroo)((k/(k+1))^k)^2 = e^-2
