Induction Problem using calculus. Let
$$
f_{o}(x)=\ln(x) \: \:\text{and} \: \:f_{n+1}(x) = \int_0^x\!f_n(t)\:\text{dt}
$$
Prove by induction that
$$
f_n(1) = - \frac{1}{n!}  \sum_{i=1}^{n} \frac{1}{i}
$$
 A: Observe that for the series $f_n(x)$:
$$
f_0(x) = \ln x \\
f_1(x) = x \ln x - x \\
f_2(x) = \frac1{2}x^2 \ln x - \frac{3x^2}{4} \\
f_3(x) = \frac1{6}x^3 \ln x - \frac{11x^3}{36}\\
\cdots \\
f_n(x) = \frac1{n!}x^n \ln x - \frac{x^n}{n!}\sum_{i=1}^n i^{-1}
$$
So we can prove this hypothesis by induction and it will encompass the desired result.
For the inductive step, use integration by parts:
$$
f_{k+1} (x) = \int_0^x \left[ \frac{t^k \ln t}{k!} - \frac{t^k }{k!} \sum_{i=1}^k \frac1{i} \right] dt \\
= \frac1{k!}\int_0^x \left[ \ln t \frac{d(\frac{t^{k+1}}{k+1})}{dt}\right]dt-\frac{x^{k+1}-0}{(k+1)!}\sum_{i=1}^k \frac1{i} \\
= \frac{x^{k+1}\ln x -0}{(k+1)!} - \frac1{(k+1)!}\int_0^x t^{k+1} \frac{d(\ln t)}{dt} dt-\frac{x^{k+1}}{(k+1)!}\sum_{i=1}^k \frac1{i}  \\
= \frac{x^{k+1}\ln x }{(k+1)!} - \frac1{(k+1)!}\int_0^x t^{k} dt-\frac{x^{k+1}}{(k+1)!}\sum_{i=1}^k \frac1{i}  \\
= \frac{x^{k+1}\ln x }{(k+1)!} - \frac{x^{k+1}}{(k+1)(k+1)!} -\frac{x^{k+1}}{(k+1)!}\sum_{i=1}^k \frac1{i}  \\
= \frac{x^{k+1}\ln x }{(k+1)!} -\frac{x^{k+1}}{(k+1)!}\sum_{i=1}^{k+1} \frac1{i}  \\
$$
From this, we can deduce that $f_n (1) = -\displaystyle\frac1{n!} \sum_{i=1}^{n} \frac1{i}$ as required.
A: Two initial cases:
$$\displaystyle n = 0: f_{0}(1) = ln(1) = 0 = -\frac{1}{0!}\sum_{i=1}^{0}\frac{1}{i}$$ This is because, that sum equals 0.
$$\displaystyle n = 1: f_{1}(1) = \int_{0}^{1}ln(t)dt = -1 = -\frac{1}{1!}\sum_{i=1}^{1}\frac{1}{i}$$
Now you have to assume that $$f_{n}(1) = -\frac{1}{n!}\sum_{i=1}^{n}\frac{1}{i}$$ and, using this fact, conclude that $$f_{n+1}(1) = -\frac{1}{(n+1)!}\sum_{i=1}^{n+1}\frac{1}{i}$$
As $f_{n+1}$ is defined based on an integral of $f_{n}$, this seems like it could be solved using Riemann sums, but I'm not that sure about it. That would be my advice for this problem.
