Convergent or divergent?

Calculus Level 3

1 − 0.9 + 2 − 1.99 + 3 − 2.999 + 4 − 3.9999 + 5 − 4.99999 + … 1-0.9+2-1.99+3-2.999+4-3.9999+5-4.99999+\ldots

For the above series ∑ k = 1 ∞ a k \displaystyle \sum_{k=1}^{\infty}a_k , we have a 2 k − 1 = k a_{2k-1}=k and a 2 k = − k + 1 0 − k a_{2k}=-k+10^{-k} .

What can we say about this series?

It converges to 1 9 \frac{1}{9} . It is divergent. It converges to some value other than 1 9 \frac{1}{9}

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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2 solutions

Brian Moehring
Nov 7, 2018

The terms a n a_n don't converge to 0 , 0, so the series diverges.

Note: This easiest way to formally conclude that the terms don't converge to 0 0 is to observe that the subsequential limit lim ⁡ k → ∞ a 2 k − 1 = lim ⁡ k → ∞ k = + ∞ \lim_{k\to\infty} a_{2k-1} = \lim_{k\to\infty} k = +\infty isn't 0. 0.

4 Helpful 0 Interesting 0 Brilliant 0 Confused
X X
Nov 7, 2018

If we put brackets like this 1 + ( − 0.9 + 2 ) + ( − 1.99 + 3 ) + ( − 2.999 + 4 ) + ( − 3.9999 + 5 ) + . . . 1+(-0.9+2)+(-1.99+3)+(-2.999+4)+(-3.9999+5)+... then it will become 1 + 1.1 + 1.01 + 1.001 + 1.0001 + . . . > 1 + 1 + 1 + 1 + 1 + . . . = ∞ 1+1.1+1.01+1.001+1.0001+...>1+1+1+1+1+...=\infty

So it is divergent.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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