Question

$\displaystyle \rm \red{\lim_{x \to \infty} \bigg (\sum\limits_{k=1}^{n} \frac{1}{k} - \displaystyle \ln \rm{(n) \bigg)}}$​

Answer 1

Answer:

For each k≥1k≥1, define

gk=1k−∫k+1kdxx

gk=1k−∫kk+1dxx

Observe that since x−1x−1 is decreasing, gk>0gk>0 for each kk. Moreover, for each kk

gk<1k−1k+1

gk<1k−1k+1

Thus, for each nn

0<∑k=1ngk<1−1n+1

0<∑k=1ngk<1−1n+1

Since each gkgk is positive,

Gn=∑k=1ngk

Gn=∑k=1ngk

is monotone increasing, moreover

lim(1−1n+1)=1

lim(1−1n+1)=1

which means GnGn is bounded above by 11. By the monotone convergence theorem, limGnlimGn exists. But

Gn=∑k=1ngk=Hn−log(n+1)

Gn=∑k=1ngk=Hn−log⁡(n+1)

Thus Hn−log(n+1)→γHn−log⁡(n+1)→γ a constant, 0<γ<10<γ<1.

NOTE Observe that the above means, since

log(1+1n)→0

log⁡(1+1n)→0

that

Hn−logn→γ

Hn−log⁡n→γ

as well.

Answer 2

Using the definition of the natural logarithm as the inverse of the exponential function, we can write:

$\displaystyle \rm \ln(n) = \int_{1}^{n} \frac{1}{x} dx$

Then, we can rewrite the limit as:

$\displaystyle \rm \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{1}{k} - \int_{1}^{n} \frac{1}{x} dx \right)$

Using the definition of the harmonic series, we can write:

$\displaystyle \rm \sum_{k=1}^{n} \frac{1}{k} = \ln(n) + \gamma + O(1/n)$

where γ is the Euler-Mascheroni constant, and the O(1/n) term represents a quantity that approaches zero as n approaches infinity.

Substituting this expression into the limit, we get:

$\displaystyle \rm \lim_{n \to \infty} \left( \ln(n) + \gamma + O(1/n) - \int_{1}^{n} \frac{1}{x} dx \right)$

Evaluating the integral, we get:

$\displaystyle \rm \int_{1}^{n} \frac{1}{x} dx = \ln(n)$

Substituting this expression into the limit, we get:

$\displaystyle \rm \lim_{n \to \infty} \left( \gamma + O(1/n) \right)$

As n approaches infinity, the O(1/n) term approaches zero, so the limit simplifies to:

$\displaystyle \rm \lim_{n \to \infty} \gamma = \gamma$

Therefore, the limit of the given expression as x approaches infinity is the Euler-Mascheroni constant γ.