integration of lim n-
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As you've shown, using integration by parts, one has
limn→∞ ∫
1
0
xnln(1+x)dx=ln(2)− limn→∞ ∫
1
0
xn+1
1+x
dx.
Hence, one we calculate the limit of the integral, we'll have the solution. Note that
|∫
1
0
xn+1
1+x
dx|≤∫
1
0
|xn+1|dx.
We have this bound since 1+x≥1, so
1
1+x
≤1. Now it is easy to see that the integral on the right-hand side has a limit of zero, and therefore
limn→∞ ∫
1
0
xnln(1+x)dx=ln(2).
limn→∞ ∫
1
0
xnln(1+x)dx=ln(2)− limn→∞ ∫
1
0
xn+1
1+x
dx.
Hence, one we calculate the limit of the integral, we'll have the solution. Note that
|∫
1
0
xn+1
1+x
dx|≤∫
1
0
|xn+1|dx.
We have this bound since 1+x≥1, so
1
1+x
≤1. Now it is easy to see that the integral on the right-hand side has a limit of zero, and therefore
limn→∞ ∫
1
0
xnln(1+x)dx=ln(2).
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