TIFR2010-Maths-A-11

Question

The series  $$\sum ^{\infty }_{n=1}\frac{(-1)^{n+1}}{\sqrt{n}}$$  

• A. Converges but not absolutely.  
• B. Converges absolutely.  
• C. Diverges.  
• D. None of the above.

Answer 1

Alternatining series test:- if series of form $\sum_{n=1}^{inf}(-1)^{n+1} a_n$  or $\sum_{n=1}^{inf}(-1)^{n} a_n$ then the series concluded to be convergent if the following conditions are meet:-

$1. a_n > a_{n+1} \forall n \in N$, consider the positive values of each term so that every term has to be greater than the succecding terms,

it means it has to be a decreasing sequence and its not neccessarily has to be decreasing sequence it can be eventually decreasing also.

 

$2. \lim_{x->\inf} a_n =0$ where $a_n $ is the general term of nth term in the sequence.

note:- for the second condition is not satisifed by the sequence then by Divergence test you can say that the given series is divergent series.

 

given series:- $\sum_{n=1}^{inf}(-1)^{n+1}\frac{1}{n^{1/2}}$

$1.$  we know that,  $n^{1/2} < (n+1)^{1/2}$,

          so ,$ \frac{1}{n^{1/2} } > \frac{1}{ (n+1)^{1/2}} \forall n \in N$, so its decreasing, the reason for decreasing is, as the first term $a_1$ is bigger than any other

term, and the second term is negative and consider its abolute value it was lesser than $a_1$,  $a_1 - a_2 > 0 $ and $a_1 - a_2 < a_1$, now whenver i was adding

$a_3$ then will increase but that doesn’t go beyond $a_1$, thats because the previously subtracted amount was greater than $a_3$, by doing like the summation

gets to value in between $0 $ and $a_1$,

$2.$ $\lim_{n->inf}(-1)^{n+1}\frac{1}{n^{1/2}}$, here we consider them as product of two limts,

->$\lim_{n->inf}(-1)^{n+1}$ * $\lim_{n->inf}\frac{1}{n^{1/2}}$ , first limit just oscillates between -1 and 1, but coming to second limit n tends infinty, it will approach value 0, so thier product is 0.

As these two conditions are satisfied by Alternating series test, we can conclude that the series converges,

Coming to Absolute convergence it will $\sum_{n=1}^{inf}\frac{1}{n^{1/2}}$, it just a p-series with value $p<1$, which implies its a divergent series.

 

so the given series, doesn’t convege absolutely and conveges by itself, which we call conditionally convergent series.

so option A is correct one.

Answer 2

By ratio test the series converges. I am not sure about absolute convergence.

Ratio test. Assume that for all n, . Suppose that there exists such that

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

 An infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute values of the summands is finite. More precisely, a real or complex series  is said to converge absolutely if for some real number . Similarly, an improper integral of a function, , is said to converge absolutely if the integral of the absolute value of the integrand is finite—that is, if 

Comments

this is not absolutely converges by $p$ series

Answer 3

The statement of Dirichlet's Test is If the sequence $\{b_n\}$ is a monotone converging to $0$ and the sequence of the partial sums $\{s_n\}$ of the series $\sum a_n$ is bounded, then the series $\sum a_nb_n$ is convergent . 

Now Since the sequence of partial sum $\{s_n\}$ of the series$\sum (-1)^{n+1}$ is bounded and the sequence $\{\frac{1}{\sqrt{n}}\}$ is monotone decreasing sequence converging to $0$. So The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ is convergent by Dirichlet's Test .

Now $\left|\frac{(-1)^{n+1}}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}$ which is divergent series by $p$ series as $0<p\le 1$