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6.29. Solution 1. We will need the following Lemma: As n → ∞, ∑ni=1 (1/i) ∼ ln(n). Proof This follows from the integral test of calculus, because !n !n n 1 (dx/x) ≤ ∑i=1 (1/i) ≤ 1 + 1 (dx/x). ! Now let us first assume that E[X1 ] = 0. Define Xi& = Xi 1{|Xi |≤i} . Also define Sn = ∑ni=1 (Xi /i) and Sn& = ∑ni=1 (Xi& /i). By the Kolmogorov maximal inequality, % & " $ n E X 2 ; |X | ≤ i & 2 & # & # 'Sn '2 1 VarSn 1 1 = ≤ . P max #Sk − E[Sk& ]# ≥ ε ln n ≤ 2 ∑ 2 2 2 2 2 ε (ln n)2 i 1≤k≤n ε ln (n) ε ln (n) i=1 Therefore, % & % & " ' n ($ 2n E X 2 ; |X | ≤ i ∞ ∞ E X 2 ; |X | ≤ i ∞ # & # 1 1 1 1 1 1 & # 2 # ≤ A P max S − E[S ] ≥ ε ln 2 ≤ A ∑ 4n ∑ 1≤k≤22n k ∑ ∑ 4n ∑ k 2 2 i i n=1 n=1 i=1 i=1 n≥log2 (i) * ) % & ∞ E X 2 ; |X | ≤ i 1 1 1 2 ≤B∑ = BE X 1 ∑ 2 ≤ CEX1 . 2 i i i=1 i≥|X | 1 By the Borel–Cantelli lemma, a.s., for all n large, ' n( # & # max n #Sk − E[Sk& ]# ≤ ε ln 22 . 1≤k≤22 n n+1 Any m is between some 22 and 22 . Therefore, ' n+1 ( # # & # # & & & #Sm − E[Sm ]# ≤ max #S − E[S ]# ≤ ε ln 22 = 2ε ln(2) · 2n ≤ 2ε ln(2) ln m. k k n+1 1≤k≤22 & − ES& |/ ln(m) → 0. But This proves that a.s., |Sm ∑k P{Sk (= Sk& } = ∑k P{|X1 | > k} < ∞ because 'X1 '1 < ∞. So m & |/ ln m → 0. But almost surely, Sk = Sk& for all k large. It suffices to prove that |ESm # # # # # # # m m # & # # E[X1 ; |X1 | ≤ i] # # E[X1 ; |X1 | > i] ## #ESm # = # ∑ # = #∑ #, #i=1 # #i=1 # i i since EX1 = 0. Thus, m # &# #ESm # ≤ ∑ E[|X1 |; |X1 | > i] . i i=1 m For all η > 0, there exists i0 such that for al i ≥ i0 , E{|X1 |; |X1 | > i} ≤ η. Therefore, ∑m i=i0 (· · · ) ≤ η ∑i=1 (1/i) ∼ i0 & |/ ln m ≤ η for all η, whence the result. η ln m. But ∑i=1 (· · · ) ≤ i0 'X1 '1 . Therefore, lim supm |ESm If µ = EX1 (= 0, then consider instead Xi◦ := Xi − µ. The preceding proves that ln−1 (n) ∑ni=1 (Xi◦ /i) → 0 a.s. Equivalently, ln−1 (n){∑ni=1 (Xi /i) − µ ∑ni=1 (1/i)} → 0. Because ∑ni=1 (1/i) ∼ ln(n), we obtain the result in general.