Question

state prove that leibnitz test​

Answer 1

Answer:

In mathematical analysis, the alternating series test is the method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion.

Formulation  :-

A series of the form

{\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}=a_{0}-a_{1}+a_{2}-a_{3}+\cdots \!}{\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}=a_{0}-a_{1}+a_{2}-a_{3}+\cdots \!}

where either all an are positive or all an are negative, is called an alternating series.

The alternating series test then says: if {\displaystyle |a_{n}|}|a_n| decreases monotonically[1] and {\displaystyle \lim _{n\to \infty }a_{n}=0}\lim _{{n\to \infty }}a_{n}=0 then the alternating series converges.

Moreover, let L denote the sum of the series, then the partial sum

{\displaystyle S_{k}=\sum _{n=0}^{k}(-1)^{n}a_{n}\!}{\displaystyle S_{k}=\sum _{n=0}^{k}(-1)^{n}a_{n}\!}

approximates L with error bounded by the next omitted term:  

{\displaystyle \left|S_{k}-L\right\vert \leq \left|S_{k}-S_{k+1}\right\vert =a_{k+1}.\!}\left | S_k - L \right \vert \le \left | S_k - S_{k+1} \right \vert = a_{k+1}.\!

proof :-

Suppose we are given a series of the form {\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1}a_{n}\!}\sum_{n=1}^\infty (-1)^{n-1} a_n\!, where {\displaystyle \lim _{n\rightarrow \infty }a_{n}=0} \lim_{n\rightarrow\infty}a_{n}=0  and {\displaystyle a_{n}\geq a_{n+1}} a_n \geq a_{n+1}  for all natural numbers n. (The case {\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}\!}\sum_{n=1}^\infty (-1)^{n} a_n\! follows by taking the negative.)[1]

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Answer 2

Answer:

Concept:

Leibnitz Theorem is basically the Leibnitz rule defined for derivative of the antiderivative. As per the rule, the derivative on nth order of the product of two functions can be expressed with the help of a formula. The functions that could probably have given function as a derivative are known as antiderivatives of the function.

Find:

state prove that leibnitz test

Given:

state prove that leibnitz test

Step-by-step explanation:

Suppose there are two functions u(t) and v(t), which have the derivatives up to nth order. Let us consider now the derivative of the product of these two functions.

The first derivative could be written as;

(uv)'=u'v+uv'

Now if we differentiate the above expression again, we get the second derivative;

(uv)"

= [(uv)']'

= (u'v+uv')'

= (u'v)'+(uv')'

=u'v+u'v' +u'v' + uv"

=u'v+2u'v' + uv"

Similarly, we can find the third derivative;

(uv)"

= [(uv)"]'

= (u"v + 2u'v' + uv")'

= (u'v)' + (2u'v')' + (uv")'

= u v + u"v' + 2u"v' + 2u'v" + u'v" + uv"

=u"v+3u"v' + 3u'v" + uv"

Now if we compare these expressions, it is found very similar to binomial expansion raised to the exponent. If we consider the terms with zero exponents, uº and vº which correspond to the functions u and v themselves, we can generate the formula for nth order of the derivative product of two functions, in a such a way that;

(uv)ⁿ = £(nt) u^(n-i) v^i

Where(nt) represents the number of i-combinations on n elements.

This formula is known as Leibniz Rule formula and can be proved by induction.

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