Which one should I use for the following problem, and how do I use it? n=1 [that capital E looking thing]
to infinity of 1/(1+ln(n))^2
to infinity of 1/(1+ln(n))^2
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Use the limit comparison test with the harmonic series:
lim(n→∞) [1/(1 + ln n)^2] / (1/n)
= lim(n→∞) n/(1 + ln n)^2
= lim(n→∞) 1/[2(1 + ln n)/n], by L'Hopital's Rule
= lim(n→∞) n/[2(1 + ln n)]
= lim(n→∞) 1/(2/n), by L'Hopital's Rule
= lim(n→∞) n/2
= ∞.
Since the harmonic series diverges, we conclude that the series in question also diverges.
I hope this helps!
lim(n→∞) [1/(1 + ln n)^2] / (1/n)
= lim(n→∞) n/(1 + ln n)^2
= lim(n→∞) 1/[2(1 + ln n)/n], by L'Hopital's Rule
= lim(n→∞) n/[2(1 + ln n)]
= lim(n→∞) 1/(2/n), by L'Hopital's Rule
= lim(n→∞) n/2
= ∞.
Since the harmonic series diverges, we conclude that the series in question also diverges.
I hope this helps!
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Use limit comparison test.
You can compare 1/(1+ln(n))^2 to 1/(n^2). As n approaches infinity, this would be 0,
Here's why you compare it to 1/(n^2),
n^2 grows exponentially as you move to the right of a graph. It grows pretty fast.
ln(n) grows slower, much slower, but it still grows, and eventually approaches infinity (just like n^2).
1/n^2 obviously would approach 0, but so would 1/ln(n)^2, or even 1/ln. Because as n approaches infinity, the denominator would continue to grow bigger, and the numerator (1) would always stay the same. Obviously, that's 0! So Yeah... It will always converge to 0.
You can compare 1/(1+ln(n))^2 to 1/(n^2). As n approaches infinity, this would be 0,
Here's why you compare it to 1/(n^2),
n^2 grows exponentially as you move to the right of a graph. It grows pretty fast.
ln(n) grows slower, much slower, but it still grows, and eventually approaches infinity (just like n^2).
1/n^2 obviously would approach 0, but so would 1/ln(n)^2, or even 1/ln. Because as n approaches infinity, the denominator would continue to grow bigger, and the numerator (1) would always stay the same. Obviously, that's 0! So Yeah... It will always converge to 0.