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Given a series:

$$\frac{1}n+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}$$

What are these types of questions called and what is the strategy for them?

The next step in the solution manual is:

$$\frac{1}{n} \left(1+\frac{1}{1+\frac{1}{n}}+\cdots+\frac{1}{1+\frac{n-1}{n}}\right)$$

And final answer for this specific question is:

$$\int_1^2 \frac{1}{x} \rightarrow \left[\ln x \vphantom{\frac 11} \right]_1^2 $$

Michael Hardy
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    The question is often to find the sum as $n$ approaches $\infty$ and the method often uses Riemann integral or bounds by inequalities. – DeepSea Jan 02 '16 at 21:26

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It's a Riemann sum. It is easy for those who have seen the definition of "Riemann sum" to be unaccustomed to recognizing them when they see them. \begin{align} & \frac{1}n+\frac{1}{n+1}+\cdots+\frac{1}{2n-1} \\[10pt] = {} & \frac 1 n \left( 1 + \frac 1 {1 + \frac 1 n} + \frac 1 {1 + \frac 2 n} + \cdots + \frac 1 {1 + \frac{n-1} n } \right) \\[10pt] = {} & \Delta x \Big( f(1) + f(1+\Delta x) + f(1+2\,\Delta x) + \cdots + f(1 + (n-1)\,\Delta x) \Big) \\[10pt] \to {} & \int_1^2 f(x)\,dx \quad\text{as }\Delta x\to 0 \qquad\text{where }f(x) = \frac 1 x. \end{align}

Notice the proper use of the $\text{“}\to\text{''}$. As I used it, it means "approaches". It can also be correctly used when $A\to B$ means $\text{“If }A\text{ then }B\text{''}$, or when $\text{“}f:A\to B\text{ ''}$ means $f$ is a function that maps the set $A$ into the set $B$. Where you used it you should have had $\text{“}=\text{''}$.

Michael Hardy
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  • Yes, you're 100% correct. While reading the definition of Riemann sums I had no idea on how to connect it to the problems. What exactly is the strategy for solving questions of this calibre? –  Jan 02 '16 at 21:36
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    @RonaldB : Probably just doing a few of them to get used to them. ${}\qquad{}$ – Michael Hardy Jan 02 '16 at 21:42
  • Am I looking for a pattern in the series? I am still confused as to what steps I have to take to solve similar problems. –  Jan 02 '16 at 21:45
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    One thing to notice is that the number of terms in the sum changes but also the individual terms change. This is not like the limit of finite partial sums of a series, where you have $a_1+\cdots+a_n$ and as $n$ grows, the number of terms changes but $a_1$ remains as it was. ${}\qquad{}$ – Michael Hardy Jan 02 '16 at 21:47