Proof of Shifting the Index of a Series of Real Numbers

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named in the step where it is cited. Let $(s_{n})_{n\in\mathbb{N}}$ and $(t_{n})_{n\in\mathbb{N}}$ be the partial sums of $(a_{k})_{k\in\mathbb{N}}$ and of $(b_{k})_{k\in\mathbb{N}}$ respectively, and let $S$ be the successor map on $\mathbb{N}$.

Step 1 (the partial sums are related by a shift). Let $n\in\mathbb{N}$. Apply Splitting a Finite Sum at an Index with the field $\mathbb{R}$, the natural numbers $1$ and $n$, and the map $k\mapsto a_{k}$ on the initial segment $[1+n]$. In the notation of that lemma the restriction $a'$ is the map on $[1]$ with $a'_{1}=a_{1}$, and $a''_{j}=a_{1+j}=b_{j}$ for every $j\in[n]$, so the lemma gives

the first summand being $a_{1}$ by claim 1 of Properties of Finite Sums. Since $1+n=n+1$ by claim 4 of Arithmetic of Addition on the Natural Numbers, this reads

Step 2 (a sequence and its one-step shift have the same limits). Let $(u_{n})_{n\in\mathbb{N}}$ be a sequence of real numbers, let $(u_{n+1})_{n\in\mathbb{N}}$ denote the sequence whose $n$-th term is $u_{n+1}$, and let $L\in\mathbb{R}$. We show that $(u_{n})_{n\in\mathbb{N}}$ converges to $L$ if and only if $(u_{n+1})_{n\in\mathbb{N}}$ converges to $L$.

Suppose first that $(u_{n})_{n\in\mathbb{N}}$ converges to $L$, and let $\varepsilon$ be a real number with $0<\varepsilon$. Choose $N\in\mathbb{N}$ with $|u_{n}-L|<\varepsilon$ for every $n\in\mathbb{N}$ with $N\le n$. Let $n\in\mathbb{N}$ with $N\le n$. By claim 6 of Properties of the Order on the Natural Numbers we have $n<n+1$, so $n\le n+1$ by claim 1 of that lemma, and a second use of claim 1 gives $N\le n+1$. Hence $|u_{n+1}-L|<\varepsilon$, and $(u_{n+1})_{n\in\mathbb{N}}$ converges to $L$.

Suppose conversely that $(u_{n+1})_{n\in\mathbb{N}}$ converges to $L$, and let $\varepsilon>0$. Choose $N\in\mathbb{N}$ with $|u_{n+1}-L|<\varepsilon$ for every $n\in\mathbb{N}$ with $N\le n$, and put $N'=N+1$. Let $m\in\mathbb{N}$ with $N'\le m$.

First, $N<m$. By claim 6 of Properties of the Order on the Natural Numbers we have $N<N+1=N'$, hence $N\le N'$ by claim 1 of that lemma, and with $N'\le m$ a second use of claim 1 gives $N\le m$. By claim 3 of that lemma exactly one of $N<m$, $N=m$, $m<N$ holds. Were $N=m$, we would have $N'\le N$; since $N<N'$ gives $N\le N'$, claim 2 of that lemma would give $N=N'$, and then $N<N$, which the first assertion of claim 2 forbids. Were $m<N$, then $m\le N$ by claim 1, and with $N\le m$ claim 2 would give $N=m$, which trichotomy excludes. Hence $N<m$.

Next, $m\ne1$. Were $m=1$, then $N<1$, so $N\le1$ by claim 1 of Properties of the Order on the Natural Numbers, while $1\le N$ by claim 4 of that lemma, so $N=1$ by claim 2; but then $N<1=N$, which the first assertion of claim 2 forbids. So $m\ne1$, and by claim 6 of Arithmetic of Addition on the Natural Numbers there is $j\in\mathbb{N}$ with $m=S(j)$; by claim 1 of that lemma $S(j)=j+1$, so $m=j+1$.

Next, $N\le j$. Suppose instead that this fails; then $j<N$ by claim 3 of Properties of the Order on the Natural Numbers, so claim 7 of that lemma provides $k\in\mathbb{N}$ with $N=j+k$. We also have $N<m$, shown above, so claim 7 provides $p\in\mathbb{N}$ with $m=N+p$. Combining, and using claim 3 of Arithmetic of Addition on the Natural Numbers,

so, rewriting both outer sums with $j$ on the right by claim 4 of Arithmetic of Addition on the Natural Numbers, we get $1+j=(k+p)+j$ and hence $1=k+p$ by claim 5 of that lemma, contradicting claim 7 of that lemma. Hence $N\le j$, and therefore $|u_{m}-L|=|u_{j+1}-L|<\varepsilon$. As $\varepsilon>0$ was arbitrary, $(u_{n})_{n\in\mathbb{N}}$ converges to $L$.

Step 3 (conclusion). Every constant sequence of real numbers converges to its value, directly from Limit of a Sequence of Real Numbers, since the absolute value of the difference is $0$ and so is smaller than every positive real number.

Suppose $\sum_{k=1}^{\infty}b_{k}$ converges, and write $T$ for its sum, so that $(t_{n})_{n\in\mathbb{N}}$ converges to $T$ by Series of Real Numbers §convergent. By claim 1 of Arithmetic of Limits of Real Sequences, applied to the constant sequence with every term equal to $a_{1}$ and to $(t_{n})_{n\in\mathbb{N}}$, the sequence $(a_{1}+t_{n})_{n\in\mathbb{N}}$ converges to $a_{1}+T$. By Step 1 that sequence is $(s_{n+1})_{n\in\mathbb{N}}$, so Step 2, applied to the sequence $(s_{n})_{n\in\mathbb{N}}$ and the number $a_{1}+T$, shows that $(s_{n})_{n\in\mathbb{N}}$ converges to $a_{1}+T$. Hence $\sum_{k=1}^{\infty}a_{k}$ converges, with sum $a_{1}+T$, which is the displayed identity.

Suppose conversely that $\sum_{k=1}^{\infty}a_{k}$ converges, and write $L$ for its sum, so that $(s_{n})_{n\in\mathbb{N}}$ converges to $L$. By Step 2 the sequence $(s_{n+1})_{n\in\mathbb{N}}$ converges to $L$, and by Step 1 its $n$-th term is $a_{1}+t_{n}$. By claim 1 of Arithmetic of Limits of Real Sequences, applied to the constant sequence with every term equal to $-a_{1}$ and to $(a_{1}+t_{n})_{n\in\mathbb{N}}$, the sequence with $n$-th term $-a_{1}+(a_{1}+t_{n})=t_{n}$ converges to $-a_{1}+L$. Hence $\sum_{k=1}^{\infty}b_{k}$ converges, with sum $-a_{1}+L$, and then $a_{1}+\sum_{k=1}^{\infty}b_{k}=a_{1}+(-a_{1}+L)=L=\sum_{k=1}^{\infty}a_{k}$, again the displayed identity.