# real- Analysis question about summation

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I don’t understand why a1+2a2+…+nan=n∑k=1kak=nSn−n−1∑k=1Ska_{1} + 2a_{2} + … + na_{n} = \sum_{k=1}^{n}ka_{k} = nS_{n} – \sum ^{n-1}_{k=1}S_{k}

I understand that nSn=na1+na2+…..nS_{n} = na_{1} + na_{2} + ….. then what is n−1∑k=1Sk\sum ^{n-1}_{k=1}S_{k}

also how limn−>∞∑n−1k=1Skn−1=a \lim_{n -> \infty} \frac{\sum ^{n-1}_{k=1}S_{k}}{n-1} = a

please help me

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– ziggurism
Oct 21 at 3:20

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2 Answers
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ziggurism has already explained what is n−1∑k=1Sk\displaystyle \sum ^{n-1}_{k=1}S_{k}.

Now, perhaps you ‘ve learnt CESARO’s theorem in that book. This is not general form of that theorem but arithmetic mean version of that theorem.
If limn→∞an=a\lim_{n \to \infty} a_n = a, then limn→∞a1+a2+a3+…+ann=a\lim_{n \to \infty} \dfrac{a_1+a_2+a_3+…+a_n}{n} = a.Here, it is the same thing (focus on SnS_n): Sn=a1+a2+a3+…+anS_n = a_1 +a_2 +a_3 +…+a_n. and ∑n−1k=1Sk=(a1)+(a1+a2)+⋯+(a1+a2+⋯+ak)+⋯+(a1+a2+⋯+an−1)=S1+S2+S3+…+Sn−1\sum^{n-1}_{k=1} S_k = (a_1) + (a_1 + a_2) + \dotsb + (a_1+a_2+\dotsb + a_k) + \dotsb + (a_1 + a_2 + \dotsb + a_{n-1})=S_1+S_2+S_3+…+S_{n-1}.

Hence, limn→∞Sn=a\lim_{n \to \infty} S_n = a, then limn→∞S1+S2+S3+…+Sn−1n−1=a\lim_{n \to \infty} \dfrac{S_1+S_2+S_3+…+S_{n-1}}{n-1} = a.
(If n→∞n \to \infty then n−1→∞n-1 \to \infty)

n−1∑k=1Sk=(a1)+(a1+a2)+⋯+(a1+a2+⋯+ak)+⋯+(a1+a2+⋯+an−1)\sum^{n-1}_{k=1} S_k = (a_1) + (a_1 + a_2) + \dotsb + (a_1+a_2+\dotsb + a_k) + \dotsb + (a_1 + a_2 + \dotsb + a_{n-1})

Count the number of times each term occurs. a1a_1 occurs n−1n-1 times. a2a_2 occurs n−2n-2 times, etc.

Therefore by associativity of addition,

n−1∑k=1Sk=(n−1)a1+(n−2)a2+⋯+an−1\sum^{n-1}_{k=1} S_k = (n-1)a_1 + (n-2)a_2 + \dotsb + a_{n-1}

Finally subtract this from nSn=na1+na2+⋯+nannS_n = na_1 + na_2 + \dotsb + na_n and you’ll have a1+2a2+⋯+nana_1 + 2a_2 + \dotsb + na_n.

and the limit asked follows from ∑n−1k=1Sk=(n−1)a1+(n−2)a2+⋯+an−1\sum^{n-1}_{k=1} S_k = (n-1)a_1 + (n-2)a_2 + \dotsb + a_{n-1} . Since 1n−1∑n−1k=1Sk=∑n−1k=1n−kn−1ak\frac{1}{n-1}\sum^{n-1}_{k=1} S_k=\sum^{n-1}_{k=1}\frac{n-k}{n-1}a_k. Using n−kn−1→1\frac{n-k}{n-1} \to 1 as n goes to infinity, the limit is ∑∞k=1ak=a\sum^{\infty}_{k=1}a_k=a.
– Fede Poncio
Oct 21 at 3:25