Otdfbt

Evaluate: $limlimits_n to inftydisplaystylesum_r=1^n dfrac12n + 2r-1$

To solve this I used the following approach:
beginalign S &= limlimits_n to inftysum_r=1^n dfracfrac 1n2 + 2frac rn-frac1n \
&= limlimits_n to inftydisplaystylesum_r=1^n dfracfrac 1n2 + 2frac rn-colorred0 = dfrac 12int _0^1 dfrac11+x,mathrmdx \ &= ln(sqrt 2)endalign

Though the answer is correct, I am unsure about my second step in which I have replaced $1/n$ by $0$. Is that allowed? I have tried it in some other problems too and it works.

"That" is not allowed in general.

It works here because you can squeeze the sum by two Riemann sums for the same function $frac12(1+x)$ on the same interval $[0,1]$:

$$frac12sum_r=1^n dfrac11 + fracrncdot frac1n = sum_r=1^n dfrac12n + 2r leq sum_r=1^n dfrac12n + 2r-1 leq sum_r=1^n dfrac12n + 2r-2 = frac12sum_r=1^n dfrac11 + fracr-1ncdot frac1n$$

No, this is not allowed, because for all you know that $frac 1 n$ could have some subtle interaction with the rest of the formula which cancels out its going to zero. For example in the limit

$$lim_nto inftynfrac1n$$

we can't just replace $frac1n$ with $0$ (yielding a limit of $0$) because the presence of the factor $n$ cancels that out. How do you know there isn't something you didn't spot in your formula which produces similar behavior? Maybe the $frac rn$ or the $frac 1n$ in the numerator has a similar "cancelling" effect on the $frac 1n$ which you're trying to replace with $0$.

By definition of Riemann integral we have $$int_0^1f(x),dx=lim_ntoinftyfrac1nsum_r=1^nf(t_r)$$ where $t_r$ lies between $(r-1)/n$ and $r/n$. For the current question the given sum matches $f(x) =dfrac12(1+x)$ and $t_r=(2r-1)/2n$. Thus the desired limit is indeed equal to $(1/2)log 2$.

The notion of a Riemann sum is far more general than usual calculus texts will lead you to believe (most calculus texts either keep $t_r=(r-1)/n$ or $t_r=r/n$). The sum in question is a Riemann sum as explained above.

Also in mathematics one can't replace $A$ by $B$ unless $A=B$. If you see some replacements which work out fine but look mysterious then you just need to understand that some steps are missing.

The technique can be used to handle sums which are not exactly Riemann sums but differ slightly from it. See this answer for an example.

I think it is better and safer to bound the limit from both sides as following:$$alpha n+2r<2n+2r-1<2n+2r$$for every $alpha<2$ and $n$ sufficiently large. Therefore$$sum_r=1^n1over 2 n+2r<sum_r=1^n1over 2n+2r-1<sum_r=1^n1over alpha n+2r$$Now fix $alpha<2$. Therefore by integrating we obtain$$1over 2ln 2letextthe desired limitle 1over 2ln2+alphaoveralpha$$Since the latter holds for every $alpha<2$, then we can write $$textthe desired limit=1over 2ln 2$$