2

$$x_n = 1 + 1/2 +\dots +1/n- \log n$$

Then -

$1.$ Is the sequence increasing?

$2.$ is the sequence convergent?

For $(1)$, $\sum 1/n$ is increasing and $\log n $ is also increasing. First few terms are increasing, but i don't know about later terms.

$(2)$ $n^{th}$ term of the sequence can be written as $a_n = (\sum_{i=1}^{n}) - \log n$

So, $\lim_{n\to \infty} a_n = \lim_{n \to \infty} \sum 1/n -\lim_{n \to \infty} \log n$

Neither first part nor second is convergent here. so i could not conclude anything.

How to solve?

Mathaddict
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2 Answers2

1

This sequences converges to the Euler–Mascheroni constant. It’s very important in number theory.

Alex J Best
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Julio Trujillo Gonzalez
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1

Let $x_n=-\log(n)+\sum_{k=1}^n \frac1k$. Then, using $\log(1+x)\ge \frac{x}{1+x}$, we see that

$$\begin{align} x_{n+1}-x_n&=\frac1{n+1}-\log\left(1+\frac1n\right)\\\\ &\le \frac1{n+1}-\frac{1}{n+1}\\\\ &=0 \end{align}$$

and $x_n$ is decreasing.

Next, we can estimate the harmonic sum as $\sum_{k=1}^n \frac1k\ge \frac12 \sum_{k=1}^{n-1}\left(\frac1k+\frac1{k+1}\right)$, which represents the Trapezoidal Rule approximation of $\int_1^n \frac1x\,dx$.

Inasmuch as $\frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $\frac1x$ and we have

$$\sum_{k=1}^n\frac1k-\log(n)\ge \sum_{k=1}^n \frac1k -\log(n)-\frac12-\frac1{2n}\ge0 $$

whence we see that

$$x_n\ge \frac12$$

Since $x_n$ is decreasing and bounded below by $\frac12$, the sequence $x_n$ converges.

Mark Viola
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