Prove that the sequence {Xn} is a divergent sequence where Xn= 1+ 1/2+ 1/3+ 1/4+.....+1/n for any natural numbers n.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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We need to prove that the series Xn is divergent where xn=1+1/2+1/3+1/4+....+1/n

 

To prove that we need to come up with another series its values equals or smaller than xn. This is called the Comparison test.

Now. let's assume the series 1+1/2+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16+....= sum yn

Now comapre it to the seires  xn

then,

1+1/2+ 1/3+1/4 + 1/5+ 1/6+1/7+1/8+ 1/9+.....+1/n = sum xn

1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16....)  = sum yn

1+1/2+1/2+1/2+1/2..... = sum yn = inf

But we observe that xn values equal or greater than yn values. Since sum yn --> inf, then sum xn-->inf

Then Xn is divergent.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To prove that 1+1/2+1/3+1/4+...+1/n is divergent.

Proof:

We write the series as below:

Let n be  natural number  such that 2^m < n < 2^(m+1).

 Xn > X2^m = 1+ 1/2 + (1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+..1/16)+(1/17+1/18+1/19......1/32)+(1/32+1/33+...1/64)+......{1/[(x^(2m-1) +1] +....1/(x^2m)}

> 1+1/2+(2/4)+(4/8)+(8/16)+x(16/32)+32/64+.....2^m-1/2^m where we took the least term in the brackets in the series in previous line and  multiplied by the number of terms in each braket. So,

Xn > X2^m = 1+1/2+1/2+1/2+1/2+1/2+1/2.... which > m/2 for 2^m number of terms. Thus S is diverging  as m--> finity(imlying n--> ifinity)

 

 

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