The expression of Euler's constant I've been given a formula for $\gamma$ which is written as follows:
$\gamma=\lim_{n\rightarrow \infty} 1+1/2+\cdots+1/n-\ln(n+1)$, which is different as the normal form of expression in terms of $\ln(n)$.
Are they the same? Is it because the limit of term $\ln(n+1)$ equals the limit of $\ln(n)$?
But theoretically that is only true when taking limits, right?
I mean if one considers sequences by themselves, $1+1/2+\cdots+1/n-\ln(n+1)$ is different as $1+1/2+\cdots+1/n-\ln(n)$?
Thanks if anyone can provide some idea.
 A: For every $n \geq 1$ we have
$$
\sum_{k=1}^{n}\frac{1}{k} - \log n = 1 - \int_{t=1}^{2}\frac{1}{t} + \frac{1}{2} - \int_{t=2}^{3}\frac{1}{t} + \cdots + \frac{1}{n-1} - \int_{t=n-1}^{n}\frac{1}{t} + \frac{1}{n};
$$
the sequences
$(\frac{1}{n})$ and
$(\int_{t=n-1}^{n}\frac{1}{t})$ are decreasing and converge to $0$;
so the convergence follows from the Leibniz's test. 
Upon noting that
$$
\sum_{k=1}^{n}\frac{1}{k} - \log (n+1) = 
1 - \int_{t=1}^{2}\frac{1}{t} + \frac{1}{2} - \int_{t=2}^{3}\frac{1}{t} + \cdots + \frac{1}{n-1} - \int_{t=n-1}^{n}\frac{1}{t} + \frac{1}{n} - \int_{t=n}^{n+1}\frac{1}{t},
$$
the same reasoning applies.
A: Both expressions are correct.  Keep in mind that
$$\log{(n+1)} = \log{n} + \log {\left (1+\frac1{n} \right )} $$
which, for large $n$, behaves as
$$\log{(n+1)} = \log{n} + \frac1{n} - \frac1{2 n^2} + \cdots$$
However, if one wanted to find
$$\lim_{n \to \infty} n \left [\sum_{k=1}^n \frac1{k} - \log{(n+1)} - \gamma  \right ] $$
it would in fact differ from
$$\lim_{n \to \infty} n \left [\sum_{k=1}^n \frac1{k} - \log{n} - \gamma  \right ] $$
A: It is incorrect to say that the limit of ln(n+1) equals the limit of ln n, because neither has a real-number limit. The def'n of Euler' constant is $$\gamma=\lim_{n\to \infty} \left(-\ln (n)+ \sum_{j=1}^{j=n}1/j\right).$$ We can write $\ln (n+k)$ instead of $\ln(n)$ in the equation above, for any constant  $k$.This changes the RHS above,for each $n$,by the amount $\ln (n+k)-\ln (n)$ which is an amount that convergess to $0$ as $n\to \infty$. (For best approximation put $k=1/2.)$ It is quite common to have functions $f,g$ where $f(n),g(n)$ go to $\infty$ as $n$ does, and yet $f(n)-g(n)$ converges.For a trivial example $f(n)=n, g(n)=n+1/n$ . Keep in mind that in many contexts ,s statement that a limit,or anything else,is equal to $\pm \infty$ is an abbreviation for a ( usually much longer) statement, not an actual equation.
