A sequence $(t_{k,n})$ st $\forall n$, $\sum_{k=1}^n t_{k,n}=1$ but $\lim_n t_{k,n} \neq 0$.

Question

I'm looking for an example of a sequence $(t_{k,n})$ such that for all $n$, $$\sum_{k=1}^n t_{k,n}=1$$ but $$\exists k: \ \lim_{n \to \infty}t_{k,n} \neq 0.$$ I've been looking at Toeplitz' lemma for sequences and I can't seem to think up an example that shows that all the assumptions are necessary. There's also another assumption, namely that $$ \sum_{k=1}^n |t_{k,n}| \mbox{ is bounded.}$$

My guesses are that these assumptions are there for the case of negative $t_{k,n}$'s - is it true that if the weights were assumed to be positive, these assumptions would follow from the first one (that they sum up to 1)?

The Toeplitz lemma says that under these assumptions on $(t_{k,n})$,

$$\lim_{n \to \infty} a_n =a \implies \lim_{n \to \infty} \sum_{k=1}^n t_{k,n} a_n =a.$$

Answer 1

How about $$t_{k,n}=\begin{cases} 1 & \text{if }k=1,\\ 0 & \text{otherwise} \end{cases}$$ so that $$\sum_{k=1}^nt_{k,n}=1+0+\cdots+0=1$$ for all $n$, and $$\lim_{n\to\infty}t_{1,n}=\lim_{n\to\infty}1=1\neq 0.$$

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You can't make an example where $t_{k,n}\neq 1$ for all $k$ and all $n$, because the requirement $$\sum_{k=1}^nt_{k,n}=1$$ forces $t_{1,1}=1$. However, you can avoid any other $1$'s, with the example of $$t_{k,n}=\begin{cases} 1 & \text{if }k=1,n=1,\\[0.1in] \tfrac{1}{2} & \text{if }k\in\{1,2\},n\geq 2,\\[0.1in] 0 & \text{otherwise} \end{cases}$$

Answer 2

$$\matrix{1&0&0&0&0&\dots\cr0&1&0&0&0&\dots\cr1&0&0&0&0&\dots\cr0&1&0&0&0&\dots\cr0&0&1&0&0&\dots\cr1&0&0&0&0&\dots\cr\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\cr}$$