Prove whether $\sum_{i=1000}^\infty a_i$, converges or diverges?

Question

Prove whether the following summation $$\sum_{i=1000}^∞ a_i$$ Where, $$ a_{2n} = \frac{-1}{2n} $$ $$ a_{2n-1} = \frac{1}{n}$$ Converges or Diverges?

Answer 1

Hint: Try writing the sum as

$$\sum_{i=1}^\infty a_i=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)=\sum_{k=1}^\infty\frac1{2k(2k-1)}$$

and comparing it to, say, the harmonic series.

The series diverges by the limit comparison test: $$\lim_{k\to\infty}\frac{\frac1{2k(2k-1)}}{\frac1k}=\frac14$$ The harmonic series diverges, so the given series does too.

Answer 2

so lets write that series as S=$\sum_{i=1}^{\infty} a_i$-$\sum_{n=1}^{1000} a_i$.because second part is finite number, let's find first sum.we can deduce that $S_i$=$a_{2*i}$+$a_{(2*i-1)}$=$1/(2*i)$.so $\sum_{i=1}^{\infty} 1/(2*i)$=$\infty$, because it is double sum of harmonic series.harmonic series is divergent, so is this problem.

Answer 3

Let $ n $ be a positive integer.

\begin{aligned}\sum_{k=1}^{2n}{a_{k}}&=\sum_{k=1}^{n}{a_{2k}}+\sum_{k=1}^{n}{a_{2k-1}}\\ &=\sum_{k=1}^{n}{-\frac{1}{2k}}+\sum_{k=1}^{n}{\frac{1}{k}}\\ \sum_{k=1}^{2n}{a_{k}}&=\frac{1}{2}\sum_{k=1}^{n}{\frac{1}{k}}\end{aligned}

Thus $ \lim\limits_{n\to +\infty}{\sum\limits_{k=1}^{2n}{a_{k}}}=+\infty $, and hence $ \left(\forall p\in\mathbb{N}^{*}\right),\ \sum\limits_{n\geq p}{a_{n}} $ DV.

Answer 4

This is a rather intuitive approach than a rigorous analytic proof of the problem.

Since,

$a_{2n-1} = \frac{1}{n}$, and $a_{2n} = \frac{-1}{2n}$, thus

we can write $a_{2n}$ in terms of $a_{2n-1}$

as $$a_{2n}=\frac{-1}{2}a_{2n-1} \tag{1}$$ From $(1)$ $$a_{1002}=\frac{-a_{1001}}{2}, a_{1004}=\frac{-a_{1003}}{2}, a_{1006}=\frac{-a_{1005}}{2} ...$$ thus, every even term can be expressed in terms of odd ones.

$$\sum_{i=1000}^{\infty} a_i= a_{1000}+a_{1001}+a_{1002}+a_{1003}...a_\infty$$

$\therefore$ $$\sum_{i=1000}^{\infty}= a_{1000}+a_{1001}-\frac{a_{1001}}{2}+a_{1003}-\frac{a_{1003}}{2} ...$$ $\therefore$ $$\sum_{i=1000}^{\infty}= a_{1000}+\frac12(a_{1001}+a_{1003}+a_{1005}+a_{1007}...a_{\infty})$$ which can be rewritten as $$\sum_{i=1000}^\infty a_i =a_{1000}+ \frac12\sum_{i=1000}^{\infty} \frac{1}{i+1}$$ which is Divergent in nature.