Let 0

Question

Let 0<α<β<1. Then lim limits n arrow ∞ displaystyle∑k=1n ∫ limits(1/(k+β))(1/(k+α)) (d x/1+x) is (A)  log e (β/α) (B)  log (1+β/1+α) (C)  log e (1+α/1+β) (D) ∞

Tardigrade Admin · Accepted Answer

lim limits n arrow ∞ displaystyle∑k=1n[ log |1+x|]1 / k+β 1 / k+α = lim limitsn arrow ∞ displaystyle∑k=1n( log (1+(1/k+α))- log (1+(1/k+β))) = lim limitsn arrow ∞ displaystyle∑k=1n( log ((k+α+1/k+α))- log ((k+β+1/k+β))) = log ((β+1/α+1))

Q. Let $0 < α < β < 1$ . Then $n \to \infty lim k = 1 \sum n \frac{1}{( k + β )} \int \frac{1}{( k + α )} \frac{d x}{1 + x}$ is

$lo g_{e} \frac{β}{α}$

$lo g \frac{1 + β}{1 + α}$

$lo g_{e} \frac{1 + α}{1 + β}$

$\infty$

Q. Let 0<α<β<1. Then n→∞lim​k=1∑n​(k+β)1​∫(k+α)1​​1+xdx​ is

loge​αβ​

log1+α1+β​

loge​1+β1+α​

∞

n→∞lim​k=1∑n​[log∣1+x∣]1/k+β​1/k+α =n→∞lim​k=1∑n​(log(1+k+α1​)−log(1+k+β1​)) =n→∞lim​k=1∑n​(log(k+αk+α+1​)−log(k+βk+β+1​)) =log(α+1β+1​)

Q. Let $0 < α < β < 1$ . Then $n \to \infty lim k = 1 \sum n \frac{1}{( k + β )} \int \frac{1}{( k + α )} \frac{d x}{1 + x}$ is

$lo g_{e} \frac{β}{α}$

$lo g \frac{1 + β}{1 + α}$

$lo g_{e} \frac{1 + α}{1 + β}$

$\infty$