Integral Inequality Question I am currently working on the problem below and I am in need of help.

Consider the definite integral $\int_{1}^{2}\frac{1}{t}dt$.
(a)By the dividing the interval $1\leq t\leq 2$ into $n$ equal parts and choosing appropriate sample points, show that
$$\sum_{j=1}^{n}\frac{1}{n+j}< \int_{1}^{2}\frac{1}{t}dt< \sum_{j=0}^{n-1}\frac{1}{n+j}$$
(b)How large should $n$ be to approximate $\int_{1}^{2}\frac{1}{t}dt$ with an error of at most $5(10^{-6}) $ using one of the sums in part (a)?
Hint: What is the difference between the underestimate and over estimate?

 A: The "appropriate" sample points are:
$1+1/n, 1+2/n,1+3/n,\ldots,2$ for the sum on the left and
$1 , 1+1/n,1+2/n,\ldots,1-(n-1)/n$ for the sum on the right.
The function $f(t)={1\over t}$ is decreasing over $[1,2]$. 
To obtain a lower estimate of the integral, we take the sum of the areas of the rectangles $$A_1, A_2, \ldots, A_n,$$ where $A_i$ is the rectangle whose base is $[1+(i-1)/n, 1+i/n]$ and whose height is $f(1+i/n)={1\over 1+{i\over n}}$ ($f$ evaluated at the right endpoint of $A_i$).  This gives the underestimate
$$
L=\sum_{i=1}^n\underbrace{ {1\over 1+{i\over n}}}_{{\rm height\ of\ }A_i }\cdot 
\underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i}.
$$
So $L$ is the sum on the left in your post.
By taking the height of $A_i$ to be $f(1+(i-1)i/n)={1\over 1+{(i-1)\over n}}$ ($f$ evaluated at the left endpoint of $A_i$), we obtain the over estimate:
$$U=\sum_{i=1}^n\underbrace{ {1\over 1+{i-1\over n}}}_{{\rm height\ of\ }A_i }\cdot 
\underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i-1}
=\sum_{i=0}^{n-1} {1\over n+i }
.
$$
So $U$ is the sum on the right in your post.
For part $b)$, using the above, if you estimate the integral with one of these sums, the error $E$ will satisfy:
$$
|E|\le U-L=\sum_{i=0}^{n-1} {1\over n+i }-\sum_{i=1}^n {1\over n+i}={1\over n}-{1\over 2n}={1\over 2n}
$$
So you just need to find the smallest value of $n$ that satisfies ${1\over 2n}\le5\cdot10^{-6}$.
A: For part (a), we divide the interval $[1,2]$ into $n$ intervals which only interests at boundary points, i.e. $[1+\frac{j-1}{n},1+\frac{j}{n}]$ where $1\leq j\leq n$. Note that $\frac{1}{t}$ is an decreasing function, which implies that for $1\leq j\leq n$
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt$$
where 
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j}{n}}dt=\frac{1}{1+\frac{j}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j}{n}}\cdot\frac{1}{n}=\frac{1}{n+j}$$
and 
$$\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{1+\frac{j-1}{n}}dt=\frac{1}{1+\frac{j-1}{n}}\cdot\Big[(1+\frac{j}{n})-(1+\frac{j-1}{n})\Big]=\frac{1}{1+\frac{j-1}{n}}\cdot\frac{1}{n}=\frac{1}{n+j-1}.$$
That is 
$$\frac{1}{n+j}<\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt<\frac{1}{n+j-1}\mbox{ for }1\leq j\leq n.$$
Now summming up the above inequalities give the result:
$$\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=1}^n\int_{1+\frac{j-1}{n}}^{1+\frac{j}{n}}\frac{1}{t}dt=\int_1^2\frac{1}{t}dt<\sum_{j=1}^n\frac{1}{n+j-1}=\sum_{j=0}^{n-1}\frac{1}{n+j}.$$
For part (b), it follows from part (a) that 
$$0<\int_1^2\frac{1}{t}dt-\sum_{j=1}^n\frac{1}{n+j}<\sum_{j=0}^{n-1}\frac{1}{n+j}-\sum_{j=1}^n\frac{1}{n+j}=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}.$$
To approximate $\int_1^2\frac{1}{t}dt$ with an error of at most $5(10^{−6})$ using the sum $\sum_{j=1}^n\frac{1}{n+j}$ in part (a), we must have
$$\frac{1}{2n}\leq 5(10^{−6}),$$
that is $n\geq 10^5$. 
