If-I-k-1-98-k-k-1-k-1-x-x-1-dx-then-1-I-gt-49-50-2-I-lt-49-50-3-I-lt-log-e-99-4-I-gt-log-e-99-

Question Number 25049 by Tinkutara last updated on 02/Dec/17

If I = \underset{k = 1}{\sum^{98}} \underset{k}{\int^{k + 1}} \frac{k + 1}{x (x + 1)} d x, then

(1) I > \frac{49}{50}

(2) I < \frac{49}{50}

(3) I < \log_{e} 99

(4) I > \log_{e} 99

Commented by ajfour last updated on 02/Dec/17

(1), (3) .

Answered by ajfour last updated on 08/Dec/17

L e t J_{k} = \int_{k}^{k + 1} \frac{k + 1}{x (x + 1)} d x

J_{k} = (k + 1) \int_{k}^{k + 1} (\frac{1}{x} - \frac{1}{x + 1}) d x

= (k + 1) \ln (\frac{x}{x + 1}) ∣_{k}^{k + 1}

= (k + 1) \ln [\frac{{(k + 1)}^{2}}{k (k + 2)}]

= 2 (k + 1) [\ln (k + 1) - \ln k - \ln (k + 2)]

= [2 (k + 1) \ln (k + 1) - k \ln k

- (k + 2) \ln (k + 2)] + [- \ln k + \ln (k + 2)]

I = \underset{k = 1}{\sum^{98}} J_{k} = 2 \underset{k = 2}{\sum^{99}} k \ln k - \underset{k = 1}{\sum^{98}} k \ln k

- \underset{k = 3}{\sum^{100}} k \ln k - \underset{k = 1}{\sum^{98}} \ln k + \underset{k = 3}{\sum^{100}} \ln k

t e r m s f r o m k = 3 t o k = 98

e n t i r e l y c a n c e l o u t, s o

I = 4 \ln 2 + 2 \times 99 \ln 99 - 2 \ln 2

- 99 \ln 99 - 100 \ln 100 - \ln 2

+ \ln 99 + \ln 100

I = \ln 2 + 99 \ln (\frac{99}{100}) + \ln 99

= \ln 99 + 99 \ln (1 - \frac{1}{100}) + \ln 2

\approx \ln 99 - 0.99 + 0.3

S o \frac{49}{50} < I < \log_{e} 99 .

Commented by Tinkutara last updated on 03/Dec/17

I a p p r e c i a t e t h i s S i r .