A mysterious series

Let a1,a2,...a100a_1,a_2,...a_{100} be real numbers each less than 11,which satisfy, a1+a2+.....a100>1a_1+a_2+.....a_{100} > 1

1.1. Let n0n_0 be the smallest integer nn such that a1+a2+.....an>1a_1+a_2+.....a_n > 1 Show that the sums an0,an0+an0−1,......,an0+.....+a1a_{n_0},a_{n_0}+a_{n_0-1} ,......,a_{n_0}+.....+a_1 are positive

2.2.Show that there exists two integers pp and qq,p<qp < q,such that the numbers aq,aq+aq−1,....,aq+.....+apa_q,a_q+a_{q-1},....,a_q+.....+a_p and ap,ap+ap+1,....,ap+.....+aqa_p,a_p+a_{p+1},....,a_p+.....+a_q

are all positive

#Sequences #Series #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

Note by Eddie The Head
7 years, 2 months ago

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Comments

For 1., we prove by contradiction. Suppose one the sums, WLOG (an0+⋯+an0−x) (a_{n_0} + \dots + a_{n_0 - x}) is not positive., i.e.(≤0) (\leq 0)

Now, As, a1+a2+⋯+an0>1 a_1 + a_2 + \dots + a_{n_0} > 1

⇒(a1+a2+⋯+an0−(x+1))+(an0−x+⋯+an0)>1 \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) + ( a_{n_0 - x} + \dots + a_{n_0}) > 1

⇒(a1+a2+⋯+an0−(x+1))>1−(an0−x+⋯+an0) \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) > 1 - ( a_{n_0 - x} + \dots + a_{n_0})

⇒(a1+a2+⋯+an0−(x+1))>1 \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) > 1 (Since (an0−x+⋯+an0) ( a_{n_0 - x} + \dots + a_{n_0}) is not positive)

But this contradicts the fact an0 a_{n_0} is the smallest n for which a1+a2+⋯+an>1 a_1 + a_2 + \dots + a_n > 1 .

Therefore our supposition is wrong and no sum is not positive, i.e. all the sums are positive

Siddhartha Srivastava - 7 years, 1 month ago
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