Let {a_n}^infinity_n = 1 be a sequence of positive real numbers. (a) Let {a_n_k}

Question

Let {a_n}^infinity_n = 1 be a sequence of positive real numbers. (a) Let {a_n_k}^infinity_k = 1 be a subsequence of {a_n}^infinity_n = 1. Prove that if {a_n}^infinity_n = 1 is convergent, then sigma^infinity_k = 1 a_n_k coverges and sigma^infinity_k = 1 a_n_k lessthanorequalto sigma^infinity_n = 1 a_n. (b) Let {a_n}^infinity_n = 1 be a decreasing sequence. Prove that if sigma^infinity_n = 1 a_n coverges, then lim_n rightarrow n na_n = 0. (c) Prove that if sigma^infinity_n = 1 a_n diverges, then sigma^infinity_n = 1 na_n also diverges.

Explanation / Answer

Note that, for every MNMN,

n=N+1ManSnn=N+1ManSM=1SNSMn=N+1ManSnn=N+1ManSM=1SNSM

hence, for every NN, considering that SM+SM+ when MM,

n=N+1anSn1.n=N+1anSn1.

This shows that the series diverges since the rest of a summable series converges to zero.

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