$\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$.

Question

How to prove the following bound: $\frac{1}{n}+\ln {n}<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$ for $n\ge2$.

My Attempt:

We have:$$\int_1^n\frac1xdx=\ln n$$So,$$\sum_{i=1}^n\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le\sum_{i=1}^n\frac1i$$

Adding $1$ on both sides to get the inequality: $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln{n}$

How Can I prove the inequality given on the LHS? Also, in this case: Why is the sum $\sum_{x=1}^n\frac{1}{x}$ more than the value of the integral $\int_1^{n}\frac{1}{x}dx$? Thanks for the help.

Answer 1

Since $e^x>1+x$ for all $x>0$, we have for $k\geq 1$ that $$ e^{\frac{1}{k}} > 1 + \frac{1}{k} = \frac{{k + 1}}{k}. $$ Thus $$ e^{\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} } > \prod\limits_{k = 1}^{n - 1} {\frac{{k + 1}}{k}} = n, $$ $$ \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} > \log n, $$ $$ \sum\limits_{k = 1}^n {\frac{1}{k}} > \log n + \frac{1}{n}. $$ To prove the upper bound note that $$ \sum\limits_{k = 1}^n {\frac{1}{k}} = 1 + \sum\limits_{k = 2}^n {\frac{1}{k}} < 1 + \sum\limits_{k = 2}^n {\int_{k - 1}^k {\frac{{dt}}{t}} } = 1 + \int_1^n {\frac{{dt}}{t}} = 1 + \log n. $$

Answer 2

Based on the comments on this solution, OP explained that the "my attempt" is more of "I saw it in my text, but wrote it incorrectly here".

As such, my writeup would not be of much help, because I thought that OP understood what they were trying in "my attempt", and just needed a pointer on what their error was.

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As indicated in my comment, your claim that

$$\sum_{i=1}^n\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le\sum_{i=1}^n\frac1i$$

is false.

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The true statement is

$$\sum_{i=1}^{n-1}\frac{1}{i+1}\le\int_1^n\frac1xdx=\ln n\le\sum_{i=1}^{n-1}\frac1i.$$

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Corollary: The desired result follows directly.

Answer 3

Proof by example:

The area under the hyperbola is $A=\log n$ and as $\dfrac1x$ is monotonous,

$$\tfrac12+\tfrac13+\tfrac14+\tfrac15+\tfrac16+\tfrac17+\tfrac18+\tfrac19+\tfrac1{10}<A<1+\tfrac12+\tfrac13+\tfrac14+\tfrac15+\tfrac16+\tfrac17+\tfrac18+\tfrac19.$$ The generalization to any $n$ is obvious.