Question

Find the values of p for which the series is \(\sum\limits_{n = 1}^\infty {\frac{{lnn}}{{{n^p}}}} \)convergent.

Short answer

The series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent for \(p > 1\) and divergent for \(p \le 1\).

Step-by-step solution

About Comparison test:-

Consider the series, \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \).

According to the comparison test, suppose that \(\sum {{a_n}} \)and \(\sum {{b_n}} \) are series with positive terms.

(i) If \(\sum {{b_n}} \)is convergent and \({a_n} \le {b_n}\forall n\)then \(\sum {{a_n}} \)is also convergent.

(ii) If \(\sum {{b_n}} \)I is divergent and \({a_n} \ge {b_n}\forall n\) then \(\sum {{a_n}} \)is also divergent.

(iii) For \(n \ge 3,\ln n > 1\) so for \(n \ge 3,\frac{{\ln n}}{{{n^p}}} > \frac{1}{{{n^p}}}\).

(iv) Apply comparison test, since only the eventual behavior of series determines convergence rather than the first several terms only.

The p-series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent if \(p > 1\) and divergent if\(p \le 1\).

Consider the case where \(p \le 1\):-

The series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) and \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) are series of positive terms. As the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) is a p-series with \(p \le 1\). Therefore, the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) is convergent.

The terms of \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) are bigger than the terms of \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \), \(\frac{{\ln n}}{{{n^p}}} > \frac{1}{{{n^p}}}\) for \(n \ge 3\).

\(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \)must also diverge. So, the series diverges for \(p \le 1\).

Consider the case where \(p > 1\):-

If \(f(x) = \frac{{\ln x}}{{{x^p}}},\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} = f(n)\).also, f is eventually decreasing. So, use the integral test. We demonstrate that f is eventually decreasing by taking its derivative withquotient rule:

\(f'(x) = \frac{{{x^p}\frac{1}{x} - \ln x.p{x^{p - 1}}}}{{{x^{2p}}}} = \frac{{{x^{p - 1}} - \ln xp{x^{p - 1}}}}{{{x^{2p}}}}\)

\( = \frac{{{x^{p - 1}}(1 - p\ln x)}}{{{x^{2p}}}}\)

The factors \({x^{p - 1}}\) and \({x^{2p}}\)are positive for positive values of x. Also \(p > 1\), for \(x \ge 3\). So, \(x \ge 3\), \(p\ln x > 1\), giving that \(0 < 1 - p\ln x\). Then for \(x \ge 3\), \(\frac{{{x^{p - 1}}(1 - p\ln x)}}{{{x^{2p}}}}\) is the product of one negative and two positive factors, therefore, negative. The integral test still applies because it is eventual behaviour of integral & series that matters for convergence, not its value from \(1 \le x \le 3\). So, f has a decreasing property that allow us to use the integral test.\(\)

Evaluation of integral:-

For evaluation of \(\int\limits_1^\infty {f(x)dx} \), first we compute the indefinite integral: \(\int {f(x)dx} = \int {\frac{{\ln x}}{{{x^p}}}dx} \).

Use integration by parts:-

Let \(u = \ln x,{\rm{ }}du = {x^{ - p}}dx\).

Then, \(du = \frac{1}{x}dx{\rm{, }}u = \frac{1}{{ - p + 1}}{x^{ - p + 1}}\).

Note that it’s important that \(p > 1\)so \( - p + 1 \ne 0\).

By the integration by parts, we have

\(\int {\frac{{\ln x}}{{{x^p}}}dx = \frac{{\ln x}}{{{x^p}}}{x^{ - p + 1}}} - \int {\frac{1}{{ - p + 1}}{x^{ - p + 1}}\frac{1}{x}dx} \)

\( = \frac{{\ln x}}{{ - p + 1}}.{x^{p + 1}} - \frac{1}{{{{( - p + 1)}^2}}}{x^{ - p + 1}}\)

\( = \frac{{\ln x - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}}\)

Take the improper integral

\(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} = \mathop {\lim }\limits_{t \to \infty } \int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \)

\( = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{\ln x - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}}} \right]_1^t\)

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){x^{p - 1}}}} + \frac{1}{{{{( - p + 1)}^2}}}\)

The term \(\ln t \to \infty \) and since \(p > 1\), also. So, the \(\frac{1}{{ - p + 1}}\) has theindefinite form\(\frac{\infty }{{ - \infty }}\) because \(\frac{1}{{ - p + 1}}\)in the numerator and \( - p + 1\) are constants.

using L’Hospital’s rule:-

Apply L’Hospital’s rule,to calculate:

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){t^{p - 1}}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{(1/t)}}{{\frac{{ - p + 1}}{p}{t^p}}}\)

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{(1/t)}}{{\frac{{ - p + 1}}{p}{t^{p + 1}}}} = 0\) [∵ p+ 1 > 2]

So, above the limit converges to zero, and thus the main limit convergent.

\( = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln t - \frac{1}{{ - p + 1}}}}{{( - p + 1){t^{p - 1}}}} + \frac{1}{{{{( - p + 1)}^2}}} = \frac{1}{{{{( - p + 1)}^2}}}\)

The integral \(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \) converges.

By integral test, \(\int\limits_1^\infty {\frac{{\ln x}}{{{x^p}}}dx} \) converges if & only if the series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \)converges. Thus, the series is convergent.

Therefore, \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^p}}}} \) is convergent for \(p > 1\) and divergent for \(p \le 1\).