Apply the first version $of$ Euler$-$Maclaurin summation formula

Rewrite ${\displaystyle \underset{1}{\overset{N}{}}}{}_1(x)f^{\text{'}}(x)dxas{\displaystyle \underset{1}{\overset{}{}}}{}_1(x)f^{\text{'}}(x)dx-{\displaystyle \underset{n}{\overset{}{}}}{}_1(x)f^{\text{'}}(x)dx$

Verify that ${\displaystyle \underset{1}{\overset{}{}}}{}_1(x)f^{\text{'}}(x)dxis$ convergent.

Identify the constant $on$ the right hand side. Note that the constant includes the

convergent integral ${\displaystyle \underset{1}{\overset{}{}}}{}_1(x)f^{\text{'}}(x)dx.$ You $do$ not need $to$ know the value $of$ this

constant, just denote $itbyA.$

For nicely behaved functions, just like $in$ our case, $we$ expect ${\displaystyle \underset{N}{\overset{}{}}}{}_1(x)f^{\text{'}}(x)dx0$

$asN.$ Now estimate this integral $asNby$ expressing $itas$

$g(N)+O(h(N))$

for some functions $g$ and $h.$

Consequently, write down $an$ asymptotic estimate $of$ the sum $as$ the final answer.

${\displaystyle \underset{k=1}{\overset{N}{}}}lnkN!{\displaystyle \underset{k=1}{\overset{N}{}}}\frac{1}{k}{H}_N{\displaystyle \underset{k=1}{\overset{N}{}}}\frac{1}{k+1}$

${\displaystyle \underset{k=1}{\overset{N}{}}}{e}^{-k}$