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?


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 $n\geq 10^5$. 
