Can we find a sequence an such that limn→∞an=∞, butlimn→∞1n∑i=1nai=a&lt;∞ ?

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Can we find a sequence an such that limn→∞an=∞, butlimn→∞1n∑i=1nai=a&amp;lt;∞ ?

Chettaf04 · Accepted Answer

Assume a locally integrable function f on [0,+∞) tends to +∞ at +∞. We will show that the antiderivative F(x)=∫0xf(t)dt satisfies  limx→+∞F(x)x=+∞  Fix C&amp;gt;0. Then there is x0&amp;gt;0 such that f(x)≥2C for all x≥x0. Thus  F(x)=∫0x0f+∫x0xf≥2C(x−x0)+∫0x0f=2Cx+D  hence  F(x)x≥2C+Dx  Now there is x1&amp;gt;x0 such that Dx≥−C for all x≥x1. Therefore  F(x)x≥2C+Dx≥2C−C=C  So we have show that Fxx tends to +∞ as claimed  Now apply this to series which fulfill your assumptions. One way to do that is to construct a piecewise constant function f which takes the value an on [n,n+1). Then f(x) behaves like an and F(x) like the partial sums.

suable9w4 · Accepted Answer

Sequence (ln⁡ln⁡n) does not satisfy your condition. No sequence can satisfy your condition if liman=+∞  Suppose the sequence (an) satisfies your condition and converges to positive infinity. Let bn=1n∑i=1nai. We shall prove that (bn) is unbounded.  Let A be an arbitrary positive number. There is an n0∈N such that for any n≥n0 we have an&amp;gt;A. Then, for any n≥n0  bn=1n(∑i=1n0ai+∑i=n0+1nai)≥A(n−n0)+constn  As n goes to ∞, the RHS converges to A, which means that for a sufficently large n we have bn&amp;gt;A−1. Since A is arbitrary, sequence (bn) is unbounded and doesnt

Can we find a sequence a_n such that \lim_{n\to\infty} a_n=\infty,

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