Determine the convergence or divergence of $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right) $.
$begingroup$
Problem
Determine the convergence or divergence of $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right). $
Solution
Notice that
$$n^{frac{1}{n^2+2n+1}}-1leq n^{frac{1}{n^2}}-1=expleft(frac{ln n}{n^2}right)-1sim frac{ln n}{n^2}.$$
Since $sumlimits_{n =1}^{infty}n^{-{frac{3}{2}}}$ is convergent, and
$$dfrac{dfrac{ln n}{n^2}}{n^{-frac{3}{2}}}=frac{ln n}{sqrt{n}}to 0(n to infty),$$
by the comparison test, $sumlimits_{n =1}^{infty}dfrac{ln n}{n^2}$ converges. Further, so does $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right)$.
proof-verification
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
$endgroup$
add a comment |
$begingroup$
Problem
Determine the convergence or divergence of $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right). $
Solution
Notice that
$$n^{frac{1}{n^2+2n+1}}-1leq n^{frac{1}{n^2}}-1=expleft(frac{ln n}{n^2}right)-1sim frac{ln n}{n^2}.$$
Since $sumlimits_{n =1}^{infty}n^{-{frac{3}{2}}}$ is convergent, and
$$dfrac{dfrac{ln n}{n^2}}{n^{-frac{3}{2}}}=frac{ln n}{sqrt{n}}to 0(n to infty),$$
by the comparison test, $sumlimits_{n =1}^{infty}dfrac{ln n}{n^2}$ converges. Further, so does $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right)$.
proof-verification
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
$endgroup$
$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48
add a comment |
$begingroup$
Problem
Determine the convergence or divergence of $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right). $
Solution
Notice that
$$n^{frac{1}{n^2+2n+1}}-1leq n^{frac{1}{n^2}}-1=expleft(frac{ln n}{n^2}right)-1sim frac{ln n}{n^2}.$$
Since $sumlimits_{n =1}^{infty}n^{-{frac{3}{2}}}$ is convergent, and
$$dfrac{dfrac{ln n}{n^2}}{n^{-frac{3}{2}}}=frac{ln n}{sqrt{n}}to 0(n to infty),$$
by the comparison test, $sumlimits_{n =1}^{infty}dfrac{ln n}{n^2}$ converges. Further, so does $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right)$.
proof-verification
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
$endgroup$
Problem
Determine the convergence or divergence of $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right). $
Solution
Notice that
$$n^{frac{1}{n^2+2n+1}}-1leq n^{frac{1}{n^2}}-1=expleft(frac{ln n}{n^2}right)-1sim frac{ln n}{n^2}.$$
Since $sumlimits_{n =1}^{infty}n^{-{frac{3}{2}}}$ is convergent, and
$$dfrac{dfrac{ln n}{n^2}}{n^{-frac{3}{2}}}=frac{ln n}{sqrt{n}}to 0(n to infty),$$
by the comparison test, $sumlimits_{n =1}^{infty}dfrac{ln n}{n^2}$ converges. Further, so does $sumlimits_{n =1}^{infty} left(n^{frac{1}{n^2+2n+1}}-1right)$.
proof-verification
proof-verification
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
edited Jan 13 at 14:48
mengdie1982
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
asked Jan 13 at 14:44
mengdie1982mengdie1982
4,912618
4,912618
$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48
add a comment |
$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48
$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48
$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48
add a comment |
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$begingroup$
This is fine once you observe the terms of the series are nonnegative.
$endgroup$
– Wojowu
Jan 13 at 14:48