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Question Number 9237 by geovane10math last updated on 25/Nov/16

A b o u t t h e E u l e r - M a s c h e r o n i c o n s t a n t

γ = \lim_{n \to \infty} (\underset{k = 1}{\sum^{n}} \frac{1}{k} - \ln n)

W h y t h e l i m i t c o n v e r g e s i f

\lim_{n \to \infty} (\underset{k = 1}{\sum^{n}} \frac{1}{k} - \ln n) = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{k} - \underset{n \to \infty}{\lim l n} n

= \infty - \infty ?

Commented by geovane10math last updated on 25/Nov/16

W h y \infty - \infty = γ ?

Commented by 123456 last updated on 25/Nov/16

this limit is idertermined, consider

\lim_{x \to \infty} x + (- x)

\lim_{x \to \infty} x = \infty

\lim_{x \to \infty} - x = - \infty

\lim_{x \to \infty} x + (- x) = \infty - \infty = ?

= \underset{x \to \infty}{\lim 0} = 0

but {can}^{'} t conclude that \infty - \infty = 0,

consider

\lim_{x \to \infty} x + (- x + 5)

\lim_{x \to \infty} - x + 5 = - \infty

\lim_{x \to \infty} x + (- x + 5) = \infty - \infty = ?

= \underset{x \to \infty}{\lim 5} = 5 ⇏ \infty - \infty = 5

by similiar way (k \in R)

\lim_{x \to \infty} x + (- x + k) = \infty - \infty

= k ⇏ \infty - \infty = k

so it depends of function nature, and do

not always diverge .

Commented by geovane10math last updated on 25/Nov/16

T h a n k s!

Commented by prakash jain last updated on 25/Nov/16

The below statement is true

if both limits exist (finite values)

\lim_{x \to a} [f (x) - g (x)] = \lim_{x \to a} f (x) - \lim_{x \to a} g (x)