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Show-that-1-1-2-1-3-1-4-1-5-1-6-is-not-convergent-Hi-

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Question Number 184918 by Mastermind last updated on 14/Jan/23
Show that 1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+... is not convergent Hi
$$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+…\:\mathrm{is}\: \\ $$$$\mathrm{not}\:\mathrm{convergent} \\ $$$$ \\ $$$$\mathrm{Hi} \\ $$
Commented by Frix last updated on 14/Jan/23
(1/5) is missing?
$$\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{is}\:\mathrm{missing}? \\ $$
Commented by Mastermind last updated on 14/Jan/23
Yes, it was a mistake ... it has been corrected tho.
$$\mathrm{Yes},\:\mathrm{it}\:\mathrm{was}\:\mathrm{a}\:\mathrm{mistake}\:…\:\mathrm{it}\:\mathrm{has}\:\mathrm{been}\: \\ $$$$\mathrm{corrected}\:\mathrm{tho}. \\ $$
Answered by Frix last updated on 14/Jan/23
Σ_(j=1) ^∞ (1/j) =Π_(p is prime ) (Σ_(k=0) ^∞ (1/p^k )) =∞ The sum (1/1)+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+... is the same as the product of these sums: (1+(1/2)+...+(1/2^n )+...)(1+(1/3)+...+(1/3^n )+...)(1+(1/5)+...+(1/5^n )+...)(1+(1/7)+...+(1/7^n )+...)...(1+(1/p)+...+(1/p^n )+...)... We know each of the sums is >1 and we know the number of primes is ∞ ...
$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{j}}\:=\underset{{p}\:\mathrm{is}\:\mathrm{prime}\:} {\prod}\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{p}^{{k}} }\right)\:=\infty \\ $$$$\mathrm{The}\:\mathrm{sum}\:\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+…\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{these}\:\mathrm{sums}: \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+…\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }+…\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}+…+\frac{\mathrm{1}}{\mathrm{5}^{{n}} }+…\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}+…+\frac{\mathrm{1}}{\mathrm{7}^{{n}} }+…\right)…\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}+…+\frac{\mathrm{1}}{{p}^{{n}} }+…\right)… \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sums}\:\mathrm{is}\:>\mathrm{1}\:\mathrm{and}\:\mathrm{we} \\ $$$$\mathrm{know}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{primes}\:\mathrm{is}\:\infty\:… \\ $$
Answered by prakash jain last updated on 14/Jan/23
1+(1/2)+((1/3)+(1/4))+((1/5)+(1/6)+(1/7)+(1/8)) +((1/9)+(1/(10))+....+(1/(16)))+... >1+(1/2)+((1/4)+(1/4))+((1/8)+(1/8)+(1/8)+(1/8))+.. =1+(1/2) (infinite times) ⇒ series is divergent
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+….+\frac{\mathrm{1}}{\mathrm{16}}\right)+… \\ $$$$>\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}\right)+.. \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\left({infinite}\:{times}\right) \\ $$$$\Rightarrow\:\mathrm{series}\:\mathrm{is}\:\mathrm{divergent} \\ $$

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