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Series

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Series And Its Convergence

Any ordered infinite sequence of terms is called a series, and is denoted by \(\sum_{i=1}^{\infty} a_{i}\).

A series \(\sum_{n-1}^{\infty} a_{n}\) converges if the sequence of the partial sums of the series converges.

The sequence of partial sums is defined as \((S_1, S_2, S_3, ..., S_n)\), where \(S_n=\sum_{k=1}^{n} a_{k} \rightarrow l\)

For example, \(\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}\) converges because \(S_{n}=1-\dfrac{1}{n+1} \rightarrow 1\). Similarly, for \(0 < x < 1\), the geometric series \(\sum_{n=0}^{\infty} x^{n}\) converges to \(\dfrac{1}{1-x}\) because \(S_{n}=\dfrac{1-x^{n+1}}{1-x}\).

A necessary condition for \(\sum_{n=1}^{\infty} a_{n}\) to converge is that \(a_{n} \rightarrow 0\). For example, for \(\vert x \vert \geq 1\), \(\sum_{n=1}^{\infty} x^{n}\) diverges because \(a_n \nrightarrow 0\). Similarly, \(\sum_{n=1}^{\infty} \sin n\) diverges because \(a_n \nrightarrow 0\).

The above condition is necessary but not sufficient. For example, \(\sum_{n=1}^{\infty} \log \left(\dfrac{n+1}{n}\right)\) diverges, however, \(\log \left(\dfrac{n+1}{n}\right) \rightarrow 0\).

Necessary and Sufficient Condition for Convergence

Theorem 1

Let \(a_{n} \geq 0 \text{ } \forall \text{ } n\). Then, \(\sum_{n=1}^{\infty} a_{n}\) converges if and only if the partial sum \(S_n\) is bounded above.

For example, the harmonic series \(\sum_{n=1}^{\infty} \dfrac{1}{n}\) diverges because

\[S_{2^{k}} \geq 1+\dfrac{1}{2}+2 \cdot \dfrac{1}{4}+4 \cdot \dfrac{1}{8}+\ldots+2^{k-1} \cdot \dfrac{1}{2^{k}}=1+\dfrac{k}{2}\]

which diverges as \(k \rightarrow \infty\).


Theorem 2

If \(\sum_{n=1}^{\infty}\vert a_{n}\vert\) converges, then \(\sum_{n=1}^{\infty} a_{n}\) converges.


Comparison Test

If \(0 \leq a_{n} \leq b_{n}\) for \(n \geq k\). Then,

(i) If \(\sum_{n=1}^{\infty} b_{n}\) converges \(\implies \sum_{n=1}^{\infty} a_{n}\) converges.

For example, \(\sum_{n=1}^{\infty} \dfrac{1}{(n)^{2}}\) converges because \(\dfrac{1}{(n)(n)} \leqslant \dfrac{1}{n(n-1)}\).

Here, \(\dfrac{1}{n(n-1)}\) is a difference series which converges to 1 as \(n \rightarrow \infty\). Therefore, \(\sum_{n=1}^{\infty} \dfrac{1}{n^{2}}\) converges.

Based on this result, we can also say that \(\sum_{n=1}^{\infty} \dfrac{1}{n !}\) converges because \(\dfrac{1}{n!} < \dfrac{1}{n^2}\) for \(n>5\).

(ii) If \(\sum_{n=1}^{\infty} a_{n}\) diverges, \(\implies \sum_{n=1}^{\infty} b_{n}\) diverges.

For example, \(\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}\) diverges because \(\dfrac{1}{n} \leqslant \dfrac{1}{\sqrt{n}}\) and the harmonic series \(\sum_{n=1}^{\infty} \dfrac{1}{n}\) diverges.


Limit Comparison Test

Let \(a_n,b_n \geq 0\) and let \(\dfrac{a_{n}}{b_{n}} \rightarrow L\). Then,

If \(L\) if positive and finite, then either both series converge or both series diverge.

Absolute And Conditional Convergence Of Series

Absolute Convergence

A series \(\sum_{t=0}^{\infty} a_{n}\) is said to converge absolutely if \(\sum_{n=0}^{\infty}\vert a_{n}\vert=L\) for some real number \(L\).


Conditional Convergence

A series \(\sum_{n-1}^{\infty} a_{n}\) converges conditionally if \(\sum_{n-1}^{\infty} a_{n}\) converges but the series \(\sum_{n-1}^{\infty}\vert a_{n}\vert\) diverges.

Rearrangement Of Series

Riemann Series Theorem: It says that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series coverges to an arbitrary real number, or diverges.


PYQs

Series And Its Convergence

1) Find the range of \(p(>0)\) for which the series:

\(\dfrac{1}{(1+a)^p}-\dfrac{1}{(2+a)^p}+\dfrac{1}{(3+a)^p}-....,a>0\), is

(i) absolutely convergent and (ii) conditionally convergent.

[2018, 10M]

Use comparison and Leibnitz tests.

2) Show that the series \(\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1}\) is conditionally convergent. (If you use any theorem, \((\mathrm{s})\) to show it, then you must give a proof of that theorem(s)).

[2016, 15M]

Use Leibnitz test

3) Test the convergence and absolute covergence of the series \(\sum_{n=1}^{\infty}(-1)^{n+1}\left(\dfrac{n}{n^{2}+1}\right)\).

[2015, 10M]

Use Leibnitz test

4) Show that the series \(\sum_{n=1}^{\infty}\left(\dfrac{\pi}{\pi+1}\right)^{n} n^{6}\) is convergent.

[2012, 12M]

Use ratio test.

5) Show that the series \(\left(\dfrac{1}{3}\right)^{2}\)+\(\left(\dfrac{1.4}{3.6}\right)^{2}\)+\(\ldots .+\left(\dfrac{1.4 .7 \ldots . .(3 n-2)}{3.6 .9 \ldots \ldots \ldots . .3 n}\right)^{2}\) till infinity, converges.

[2009, 15M]

Use Rabee's test.

6) Find all the positive values of \(a\) for which the series \(\sum_{n=1}^{\infty} \dfrac{(a n)^{n}}{n !}\) converges.

[2002, 12M]

Use ratio test.

Rearrangement of Series

1) Let \(\sum_{n=1}^{\infty} x_{n}\) be a conditionally convergent series of real numbers. Show that there is a rearrangement \(\sum_{n=1}^{\infty} x_{\pi(n)}\) of the series \(\sum_{n=1}^{\infty} x_{n}\) that converges to 100.

[2017, 20M]


2) Show that the series \(\sum \dfrac{1}{n(n+1)}\) is equivalent to \(\dfrac{1}{2} \prod_{2}^{\infty}\left(1+\dfrac{1}{n^{2}-1}\right)\).

[2008, 15M]


3) Rearrange the series \(\sum_{n=1}^{\infty}(-1)^{n+1} \dfrac{1}{n}\) to converge to 1.

[2007, 20M]


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