Problem involving harmonic number The problem is as follows and I have trouble solving part $(b)$:


I attempted the problem in the following way:
Let $d(i)$ be the total length covered before the end of $i$th second. Then $d(i)=d(i-1)*\frac{i}{i-1}+1$. This got simplified to 
$$d(i)=\frac{i}{i-(i-1)}+\frac{i}{i-(i-2)}+\dots +\frac{i}{i-1}+1$$
Thus $d(i)=i(\frac{1}{1}+\frac{1}{2}+\dots +\frac{1}{i-1})+1$ and hence 
$$d(i)=iH_{i-1}+1$$
Now length of rug during $i$th second is $100i$, and thus fraction of rug crossed over first $i$ seconds is $d(i)/100i$. And this was very difficult to solve in part $(c)$.
However the solution states that fraction of rug crossed during $i$th second is $\frac{1}{100i}$, therefore fraction of rug crossed altogether over $n$ seconds is $\sum_{i=1}^{n}\frac{1}{100i}$. When I checked this is indeed equal to my solution of $d(i)/100i$, however I can't figure out how this solution is correct. Can someone help me understand it and realize how I should have approached it in this(better) way.
EDIT:$d(1)=1$ as distance covered during $1$st second is $1$ cm
 A: You aren't really answering the questions in the order they are being asked, and if you did, you would quickly realize that dealing with the distance function is not as convenient as dealing with the fraction function.
Part (a) specifically asks for the fraction $f(i)$ of the total rug's length that the bug covers in the time interval $t \in (i-1, i]$.  It is important to observe that this fraction does not depend on whether we measure it at the moment $t = i$ before the rug is stretched, or after it is stretched but the bug has not yet begun to crawl for the next time interval $t \in (i, i+1]$.  We can immediately see this is true because the rug is stretched uniformly, thus the fraction of distance traveled in $t \in (i-1,i]$ is also scaled by the same factor that the rug is scaled.
Next, the bug always crawls $1$ cm in each interval $t \in (i-1, i]$, regardless of $i$ (here, $i$ is always a positive integer).  So, the key to answering (a) is to note that the fraction of the rug that the bug crawls in this interval is simply a function of the total length of the rug during this time, which is of course determined by the number of times it has been stretched.  Since in the first interval $t \in (0,1]$ the rug is $100$ cm, and it is lengthened by $100$ cm in each successive interval, we find that the answer to (a) is $$f(i) = \frac{1}{100i},$$
and note this is a dimensionless quantity.
Part (b) asks us to find the cumulative fraction of the distance traveled, which we will call $c(n)$, over $t \in (0, n]$ for positive integers $n$.  Again, this is simple; it is just the sum of the individual fractions traveled:  $$c(n) = \sum_{i=1}^n f(i) = \frac{H_n}{100},$$ where $H_n$ is the $n^{\rm th}$ harmonic number.  This is again because the stretching action operates uniformly:  if the bug has cumulatively traveled some fraction $x \in [0,1]$ of the rug at time $t$, this is true regardless of the length of the rug.  Even if we stretch it, it still has traveled this fraction of the rug's length.  Then, when the bug traverses the next fraction, these fractions add to get the next cumulative fraction traveled.
(If the above still isn't convincing, consider a small numerical example.  Suppose the bug is at $3$ cm of a $5$ cm long rug; that is, it is $3/5^{\rm th}$ of the way across.  Then over the next time increment, it travels $1$ cm, reaching $4$ cm and then you stretch the rug $1$ cm to make it $6$ cm.  Now because the fraction traversed is $1/5$ (it traveled $1$ when the rug was $5$), but it was already $3/5$ the way across, it is now $3/5 + 1/5 = 4/5$ the way across.  This is verified by noting that before you stretched it, the bug was at $4$ cm for a $5$ cm rug, and the stretch to $6$ cm doesn't change the fraction--after the stretch the bug is at $4(6/5) = 4.8$ cm.)
For the final part, it becomes obvious that you need to find the smallest integer $n$ such that $c(n) \ge 1$; i.e., $H_n \ge 100$.  This is a straightforward exercise in calculus, which I leave to the reader.  As a final hint, consider a suitable approximation of the sum $H_n$ by an integral, and thereby obtain a function that is bounded below by $H_{n-1}$ and above by $H_n$.

As an aside, it may be worth noting that using your notation, we simply have $$d(i) = i H_{i-1} + 1 = i\left(H_{i-1} + \frac{1}{i}\right) = i H_i,$$ thus $$\frac{d(n)}{100n} = \frac{H_n}{100}.$$
